NCERT Solutions · Class 9 Mathematics I’m Up and Down, and Round and Round

Two of its sides are radii of the same circle.Let the circle have centre C and radius \(\displaystyle r\), and let AB be a chord of it. Joining the two ends of the chord to the centre gives \(\displaystyle \triangle \mathrm{CAB}\).A and B are points on the circle. By the very definition of a circle — the set of points at a fixed distance from the centre — every point of the circle is at distance \(\displaystyle r\) from C. So \[\mathrm{CA}=r \quad\text{and}\quad \mathrm{CB}=r, \qquad\text{hence}\qquad \mathrm{CA}=\mathrm{CB}. \]A triangle with two equal sides is isosceles.Because the two equal sides are CA and CB, the base is the chord AB, and the base angles are equal: \[\angle \mathrm{CAB}=\angle \mathrm{CBA}. \]One thing to be careful about: if the chord AB happens to be a diameter, then C lies on AB, the three points A, C, B are collinear and there is no triangle at all. So the statement is about a chord that is not a diameter.Every chord of a circle, together with the centre, forms an isosceles triangle — the two radii are its equal sides and the chord is its base.

SSS congruence — the four radii give two pairs of equal sides for free.Let AB and DE be two chords of the same circle, with centre C and radius \(\displaystyle r\). By the previous question, \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\) are both isosceles, with bases AB and DE. "Equal base length" means \[\mathrm{AB}=\mathrm{DE}. \]Now compare \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\) side by side:

In \(\displaystyle \triangle \mathrm{CAB}\) and \(\displaystyle \triangle \mathrm{CDE}\)	Reason
\(\displaystyle \mathrm{CA}=\mathrm{CD}\)	both are radii, each equal to \(\displaystyle r\)
\(\displaystyle \mathrm{CB}=\mathrm{CE}\)	both are radii, each equal to \(\displaystyle r\)
\(\displaystyle \mathrm{AB}=\mathrm{DE}\)	given (equal bases)

All three pairs of corresponding sides are equal, so by the SSS congruence criterion, \[\triangle \mathrm{CAB}\cong\triangle \mathrm{CDE}. \]Since corresponding parts of congruent triangles are equal, this also gives \(\displaystyle \angle \mathrm{ACB}=\angle \mathrm{DCE}\) — which is exactly Theorem $\displaystyle 2$, that equal chords subtend equal angles at the centre.A word of caution: the argument uses \(\displaystyle \mathrm{CA}=\mathrm{CD}\) and \(\displaystyle \mathrm{CB}=\mathrm{CE}\), and that needs the two chords to belong to the same circle (or to two circles of the same radius). Equal chords in circles of different sizes give triangles that are certainly not congruent.The two isosceles triangles are congruent, by SSS.

RHS congruence — and then a second proof using Baudhāyana–Pythagoras.The set-up (Fig. $\displaystyle 5.12$): a circle with centre C and a chord AB. The perpendicular from C meets AB at M, so we are given \(\displaystyle \angle \mathrm{CMA}=\angle \mathrm{CMB}=90^\circ\). We must show \(\displaystyle \mathrm{AM}=\mathrm{BM}\).Notice how this reverses Theorem 4. There we were told M was the midpoint and deduced the right angle; here we are told the right angle and must deduce the midpoint. That is what "converse" means.First proof (congruence). Compare \(\displaystyle \triangle \mathrm{CMA}\) and \(\displaystyle \triangle \mathrm{CMB}\):

Statement	Reason
\(\displaystyle \angle \mathrm{CMA}=\angle \mathrm{CMB}=90^\circ\)	given
\(\displaystyle \mathrm{CA}=\mathrm{CB}\)	both are radii (the hypotenuses)
\(\displaystyle \mathrm{CM}=\mathrm{CM}\)	common side

A right angle, an equal hypotenuse and an equal side: by the RHS congruence criterion, \(\displaystyle \triangle \mathrm{CMA}\cong\triangle \mathrm{CMB}\). Corresponding sides of congruent triangles are equal, so \(\displaystyle \mathrm{AM}=\mathrm{BM}\).(Note we could not have used SAS here: the equal angle at M is not the angle between the two sides CM and CA. RHS is the criterion built for exactly this situation.)Second proof (Baudhāyana–Pythagoras). Both triangles are right-angled at M, and \(\displaystyle \mathrm{CA}=\mathrm{CB}=r\). So \[\mathrm{AM}^{2}=\mathrm{CA}^{2}-\mathrm{CM}^{2}=r^{2}-\mathrm{CM}^{2}, \qquad \mathrm{BM}^{2}=\mathrm{CB}^{2}-\mathrm{CM}^{2}=r^{2}-\mathrm{CM}^{2}. \] The two right-hand sides are the same number, so \(\displaystyle \mathrm{AM}^{2}=\mathrm{BM}^{2}\). Lengths are positive, so \(\displaystyle \mathrm{AM}=\mathrm{BM}\).So M is the midpoint of AB: the perpendicular from the centre of a circle to a chord bisects the chord. This is Theorem 5.

Both A and the centre are equidistant from B and C — so both sit on the perpendicular bisector of BC.Let the circle have centre O, and let \(\displaystyle \triangle \mathrm{ABC}\) be inscribed in it with \(\displaystyle \mathrm{AB}=\mathrm{AC}\). Let \(\displaystyle \ell\) denote the perpendicular bisector of BC.Step $\displaystyle 1$: A lies on \(\displaystyle \ell\). We are given \(\displaystyle \mathrm{AB}=\mathrm{AC}\), so A is equidistant from B and C. The perpendicular bisector of BC is precisely the locus of points equidistant from B and C (Section $\displaystyle 5.2$). Hence A lies on \(\displaystyle \ell\).Step $\displaystyle 2$: O lies on \(\displaystyle \ell\). B and C are points of the circle, so \(\displaystyle \mathrm{OB}=\mathrm{OC}=\) the radius. So O too is equidistant from B and C, and therefore O also lies on \(\displaystyle \ell\).Step $\displaystyle 3$: \(\displaystyle \ell\) is the altitude from A. By definition \(\displaystyle \ell\) is perpendicular to BC, and by Step $\displaystyle 1$ it passes through A. Now, through a given point there is exactly one line perpendicular to a given line. The altitude from A to BC is, by definition, the line through A perpendicular to BC. Since \(\displaystyle \ell\) is a line through A perpendicular to BC, it must be that very line: \[\text{altitude from A to BC} \;=\; \ell. \]Step $\displaystyle 4$: finish. By Step $\displaystyle 2$, O lies on \(\displaystyle \ell\), and by Step $\displaystyle 3$, \(\displaystyle \ell\) is the altitude from A. So the altitude passes through O.(A is not on the line BC, since A, B, C are vertices of a triangle, so "the altitude from A" and Step $\displaystyle 1$ both make sense.)Worth noticing: in this isosceles triangle the altitude from A, the median from A, the perpendicular bisector of BC and the bisector of \(\displaystyle \angle \mathrm{BAC}\) are all the same line — and that single line contains the centre of the circle. This is the line of reflection symmetry of the whole figure.Hence the altitude from A to BC passes through the centre of the circle.

Find each chord's distance from the centre by Baudhāyana–Pythagoras, then add — because both midpoints lie on one line through the centre.Let the circle have centre O and radius \(\displaystyle r=5\ \mathrm{cm}\). Let \(\displaystyle \mathrm{AB}=8\ \mathrm{cm}\) with midpoint P, and \(\displaystyle \mathrm{CD}=6\ \mathrm{cm}\) with midpoint Q, the two chords being parallel and on opposite sides of O.Step $\displaystyle 1$ — the perpendiculars. By Theorem $\displaystyle 4$, the line joining the centre to the midpoint of a chord is perpendicular to that chord. So \(\displaystyle \mathrm{OP}\perp \mathrm{AB}\) and \(\displaystyle \mathrm{OQ}\perp \mathrm{CD}\), which means OP and OQ are the distances of the two chords from the centre.Step $\displaystyle 2$ — the $\displaystyle 8$ cm chord. P is the midpoint, so \(\displaystyle \mathrm{AP}=4\ \mathrm{cm}\). In \(\displaystyle \triangle \mathrm{OPA}\), right-angled at P, with hypotenuse \(\displaystyle \mathrm{OA}=5\ \mathrm{cm}\): \[\mathrm{OP}=\sqrt{5^{2}-4^{2}}=\sqrt{25-16}=\sqrt{9}=3\ \mathrm{cm}. \]Step $\displaystyle 3$ — the $\displaystyle 6$ cm chord. Q is the midpoint, so \(\displaystyle \mathrm{CQ}=3\ \mathrm{cm}\). In \(\displaystyle \triangle \mathrm{OQC}\), right-angled at Q, with hypotenuse \(\displaystyle \mathrm{OC}=5\ \mathrm{cm}\): \[\mathrm{OQ}=\sqrt{5^{2}-3^{2}}=\sqrt{25-9}=\sqrt{16}=4\ \mathrm{cm}. \](Sanity check against Theorem $\displaystyle 8$: the longer chord, $\displaystyle 8$ cm, came out closer to the centre, $\displaystyle 3$ cm against $\displaystyle 4$ cm. Good.)Step $\displaystyle 4$ — why P, O, Q are on one straight line. OP is perpendicular to AB and OQ is perpendicular to CD; but \(\displaystyle \mathrm{AB}\parallel \mathrm{CD}\), so OQ is perpendicular to the direction of AB as well. Through the single point O there is exactly one line perpendicular to that direction. So OP and OQ lie along the same line, and P, O, Q are collinear.Step $\displaystyle 5$ — add or subtract? The chords are on opposite sides of the centre, so P and Q lie on opposite sides of O on that common line. Therefore the two distances add: \[\mathrm{PQ}=\mathrm{OP}+\mathrm{OQ}=3+4=7\ \mathrm{cm}. \](Had the chords been on the same side of the centre, P and Q would be on the same side of O and we would subtract, giving \(\displaystyle 4-3=1\ \mathrm{cm}\). The phrase "on opposite sides" in the question is doing real work.)The distance between the midpoints of the chords is $\displaystyle 7$ cm.

Equal chords have equal halves; the radius is the same hypotenuse both times.Theorem $\displaystyle 6$ says: chords of a circle having the same length are all at the same distance from the centre.Take a circle with centre C and radius \(\displaystyle r\), and two chords \(\displaystyle \mathrm{AB}\) and \(\displaystyle \mathrm{FG}\) with \(\displaystyle \mathrm{AB}=\mathrm{FG}\) (Fig. $\displaystyle 5.14$). Let E and H be the feet of the perpendiculars dropped from C onto AB and FG. Then CE and CH are exactly the distances of the two chords from the centre, and what we must show is \(\displaystyle \mathrm{CE}=\mathrm{CH}\).Step $\displaystyle 1$ — halve the chords. By Theorem $\displaystyle 5$, the perpendicular from the centre to a chord bisects it. So E is the midpoint of AB and H is the midpoint of FG: \[\mathrm{AE}=\tfrac{1}{2}\mathrm{AB}, \qquad \mathrm{FH}=\tfrac{1}{2}\mathrm{FG}. \] Since \(\displaystyle \mathrm{AB}=\mathrm{FG}\), halving both gives \[\mathrm{AE}=\mathrm{FH}. \]Step $\displaystyle 2$ — apply Baudhāyana–Pythagoras twice. \(\displaystyle \triangle \mathrm{CEA}\) is right-angled at E, with hypotenuse \(\displaystyle \mathrm{CA}=r\): \[\mathrm{CA}^{2}=\mathrm{CE}^{2}+\mathrm{AE}^{2} \quad\Longrightarrow\quad \mathrm{CE}^{2}=r^{2}-\mathrm{AE}^{2}. \] \(\displaystyle \triangle \mathrm{CHF}\) is right-angled at H, with hypotenuse \(\displaystyle \mathrm{CF}=r\): \[\mathrm{CF}^{2}=\mathrm{CH}^{2}+\mathrm{FH}^{2} \quad\Longrightarrow\quad \mathrm{CH}^{2}=r^{2}-\mathrm{FH}^{2}. \]Step $\displaystyle 3$ — compare. By Step $\displaystyle 1$, \(\displaystyle \mathrm{AE}=\mathrm{FH}\), so \(\displaystyle \mathrm{AE}^{2}=\mathrm{FH}^{2}\), so the two right-hand sides \(\displaystyle r^{2}-\mathrm{AE}^{2}\) and \(\displaystyle r^{2}-\mathrm{FH}^{2}\) are the same number. Hence \[\mathrm{CE}^{2}=\mathrm{CH}^{2}, \] and since distances are positive, \(\displaystyle \mathrm{CE}=\mathrm{CH}\).Notice what makes this work: the two right triangles have the same hypotenuse length (the radius) and the same other leg (half the chord), so the remaining leg has no choice but to match.Chords of equal length are equidistant from the centre — Theorem 6.

RHS congruence on the two right triangles \(\displaystyle \triangle \mathrm{CEA}\) and \(\displaystyle \triangle \mathrm{CHF}\).In Fig. $\displaystyle 5.15$, C is the centre of the circle, AB and GF are two chords, CE is the perpendicular from C to AB (with E on AB), and CH is the perpendicular from C to the chord GF (with H on it). We are given \(\displaystyle \mathrm{CE}=\mathrm{CH}\), so the two chords are the same distance from the centre. We must show \(\displaystyle \mathrm{AB}=\mathrm{GF}\). This is the converse of Theorem 6.Step $\displaystyle 1$ — the feet are midpoints. By Theorem $\displaystyle 5$, the perpendicular from the centre to a chord bisects the chord. So \[\mathrm{AB}=2\,\mathrm{AE} \qquad\text{and}\qquad \mathrm{GF}=2\,\mathrm{FH}. \] So it is enough to prove \(\displaystyle \mathrm{AE}=\mathrm{FH}\).Step $\displaystyle 2$ — compare the two right triangles.

In \(\displaystyle \triangle \mathrm{CEA}\) and \(\displaystyle \triangle \mathrm{CHF}\)	Reason
\(\displaystyle \angle \mathrm{CEA}=\angle \mathrm{CHF}=90^\circ\)	CE \(\displaystyle \perp\) AB and CH \(\displaystyle \perp\) GF
\(\displaystyle \mathrm{CA}=\mathrm{CF}\)	both are radii — the hypotenuses
\(\displaystyle \mathrm{CE}=\mathrm{CH}\)	given

Right angle, hypotenuse, side: by the RHS congruence criterion, \[\triangle \mathrm{CEA}\cong\triangle \mathrm{CHF}. \]Step $\displaystyle 3$ — finish. Corresponding sides of congruent triangles are equal, so \(\displaystyle \mathrm{AE}=\mathrm{FH}\). Doubling both sides and using Step $\displaystyle 1$, \[\mathrm{AB}=2\,\mathrm{AE}=2\,\mathrm{FH}=\mathrm{GF}. \]\(\displaystyle \mathrm{AB}=\mathrm{GF}\): chords of a circle that are equidistant from the centre have equal length — this is Theorem 7.

Same figure, but replace the congruence step with Baudhāyana–Pythagoras.Work again in Fig. $\displaystyle 5.15$: circle with centre C and radius \(\displaystyle r\); \(\displaystyle \mathrm{CE}\perp \mathrm{AB}\) with E on AB, \(\displaystyle \mathrm{CH}\perp \mathrm{GF}\) with H on GF, and \(\displaystyle \mathrm{CE}=\mathrm{CH}\).Step $\displaystyle 1$ — the feet are midpoints. By Theorem $\displaystyle 5$ the perpendicular from the centre bisects the chord, so \(\displaystyle \mathrm{AB}=2\,\mathrm{AE}\) and \(\displaystyle \mathrm{GF}=2\,\mathrm{FH}\).Step $\displaystyle 2$ — write each half-chord in terms of the radius and the distance. \(\displaystyle \triangle \mathrm{CEA}\) is right-angled at E, and its hypotenuse CA is a radius, so \[\mathrm{AE}^{2}=\mathrm{CA}^{2}-\mathrm{CE}^{2}=r^{2}-\mathrm{CE}^{2}. \] \(\displaystyle \triangle \mathrm{CHF}\) is right-angled at H, and its hypotenuse CF is a radius, so \[\mathrm{FH}^{2}=\mathrm{CF}^{2}-\mathrm{CH}^{2}=r^{2}-\mathrm{CH}^{2}. \]Step $\displaystyle 3$ — use the given equality. We are told \(\displaystyle \mathrm{CE}=\mathrm{CH}\), so \(\displaystyle \mathrm{CE}^{2}=\mathrm{CH}^{2}\), and therefore \[r^{2}-\mathrm{CE}^{2}=r^{2}-\mathrm{CH}^{2} \quad\Longrightarrow\quad \mathrm{AE}^{2}=\mathrm{FH}^{2} \quad\Longrightarrow\quad \mathrm{AE}=\mathrm{FH}, \] the last step because lengths are positive.Step $\displaystyle 4$ — double. \[\mathrm{AB}=2\,\mathrm{AE}=2\,\mathrm{FH}=\mathrm{GF}. \]Compare the two methods: the congruence proof of the previous question and this one do the same work. RHS congruence is really just Baudhāyana–Pythagoras in disguise — it is exactly the statement that in a right triangle the hypotenuse and one leg determine the third leg.\(\displaystyle \mathrm{AB}=\mathrm{GF}\).

Half the chord, the distance, and the radius form a right triangle.Let the circle have centre O and radius \(\displaystyle r=7\ \mathrm{cm}\), let AB be the chord, and let M be the foot of the perpendicular from O to AB, so \(\displaystyle \mathrm{OM}=6\ \mathrm{cm}\).First check such a chord can exist: the distance \(\displaystyle 6\ \mathrm{cm}\) is less than the radius \(\displaystyle 7\ \mathrm{cm}\), so the line does cut the circle.Step $\displaystyle 1$ — M is the midpoint. By Theorem $\displaystyle 5$, the perpendicular from the centre to a chord bisects it, so \(\displaystyle \mathrm{AM}=\tfrac{1}{2}\mathrm{AB}\).Step $\displaystyle 2$ — the right triangle. \(\displaystyle \triangle \mathrm{OMA}\) is right-angled at M. Its hypotenuse is \(\displaystyle \mathrm{OA}=r=7\ \mathrm{cm}\) (a radius, since A is on the circle), and its legs are \(\displaystyle \mathrm{OM}=6\ \mathrm{cm}\) and AM. By the Baudhāyana–Pythagoras theorem, \[\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}, \] \[7^{2}=6^{2}+\mathrm{AM}^{2}, \] \[\mathrm{AM}^{2}=49-36=13, \qquad \mathrm{AM}=\sqrt{13}\ \mathrm{cm}. \]Step $\displaystyle 3$ — double it. \[\mathrm{AB}=2\,\mathrm{AM}=2\sqrt{13}\ \mathrm{cm}. \]Numerically \(\displaystyle \sqrt{13}\approx 3.606\), so \(\displaystyle \mathrm{AB}\approx 7.21\ \mathrm{cm}\).A quick plausibility check: the longest possible chord here is the diameter, \(\displaystyle 14\ \mathrm{cm}\), and this chord sits far out from the centre, so a value of about \(\displaystyle 7.2\ \mathrm{cm}\) — roughly half the diameter — is sensible.The chord is \(\displaystyle 2\sqrt{13}\ \mathrm{cm}\approx 7.2\ \mathrm{cm}\) long.

It is the previous question done with letters instead of numbers.Let the circle have centre O and radius \(\displaystyle r\), let AB be a chord, and let M be the foot of the perpendicular from O to AB, so the perpendicular distance of the chord from the centre is \(\displaystyle \mathrm{OM}=d\).Step $\displaystyle 1$ — M is the midpoint of AB. This is Theorem $\displaystyle 5$: the perpendicular from the centre of a circle to a chord bisects the chord. So \[\mathrm{AM}=\tfrac{1}{2}\mathrm{AB}, \qquad\text{that is,}\qquad \mathrm{AB}=2\,\mathrm{AM}. \] This step is the whole reason a single formula can work: it converts "the chord" into "twice one particular leg of a right triangle".Step $\displaystyle 2$ — build the right triangle. Join OA. Since A lies on the circle, \(\displaystyle \mathrm{OA}=r\). And \(\displaystyle \angle \mathrm{OMA}=90^\circ\), because OM is perpendicular to AB. So \(\displaystyle \triangle \mathrm{OMA}\) is right-angled at M, with hypotenuse OA.Step $\displaystyle 3$ — Baudhāyana–Pythagoras. \[\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}, \] \[r^{2}=d^{2}+\mathrm{AM}^{2}, \] \[\mathrm{AM}^{2}=r^{2}-d^{2}, \qquad \mathrm{AM}=\sqrt{r^{2}-d^{2}} \] (taking the positive square root, since AM is a length).Step $\displaystyle 4$ — double it. \[\mathrm{AB}=2\,\mathrm{AM}=2\sqrt{r^{2}-d^{2}}. \]Does the formula behave sensibly? Three checks: The formula also re-proves Theorem $\displaystyle 6$ and Theorem $\displaystyle 7$ in one line: for a fixed \(\displaystyle r\), the chord length depends on \(\displaystyle d\) alone, so equal \(\displaystyle d\) forces equal chord and equal chord forces equal \(\displaystyle d\). And since \(\displaystyle 2\sqrt{r^{2}-d^{2}}\) gets smaller as \(\displaystyle d\) gets bigger, it re-proves Theorem $\displaystyle 8$ as well.Hence the chord length is \(\displaystyle 2\sqrt{r^{2}-d^{2}}\).

No. Test the claim with the formula \(\displaystyle 2\sqrt{r^{2}-d^{2}}\) — the relationship is not a proportion.Write \(\displaystyle r\) for the radius and let CD be at distance \(\displaystyle d\) from the centre, so AB is at distance \(\displaystyle 2d\). By the formula from Question $\displaystyle 2$, \[\mathrm{CD}=2\sqrt{r^{2}-d^{2}}, \qquad \mathrm{AB}=2\sqrt{r^{2}-4d^{2}}. \] (For AB to exist at all we need \(\displaystyle 2d<r\).)A counterexample settles it. Take \(\displaystyle r=5\ \mathrm{cm}\) and \(\displaystyle d=1\ \mathrm{cm}\), so CD is \(\displaystyle 1\ \mathrm{cm}\) from the centre and AB is \(\displaystyle 2\ \mathrm{cm}\) from the centre: \[\mathrm{CD}=2\sqrt{25-1}=2\sqrt{24}\approx 9.80\ \mathrm{cm}, \qquad \mathrm{AB}=2\sqrt{25-4}=2\sqrt{21}\approx 9.17\ \mathrm{cm}. \] CD is longer than AB, but only by about \(\displaystyle 7\%\) — nowhere near twice. So \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) does not follow.Why the intuition fails. Doubling \(\displaystyle d\) does not double or halve anything inside the formula, because \(\displaystyle d\) enters as \(\displaystyle d^{2}\) and then under a square root, subtracted from \(\displaystyle r^{2}\). The chord length is not proportional to the distance, nor inversely proportional to it — the two are not in any fixed ratio at all.What we can say. AB is farther from the centre than CD, so by Theorem $\displaystyle 8$ the farther chord is the shorter one: \[\mathrm{CD}>\mathrm{AB}. \] That inequality is all that follows from the given information.When does \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) actually happen? Solve for it: \[2\sqrt{r^{2}-d^{2}}=2\cdot 2\sqrt{r^{2}-4d^{2}} \] \[\sqrt{r^{2}-d^{2}}=2\sqrt{r^{2}-4d^{2}} \] Squaring both sides, \[r^{2}-d^{2}=4\left(r^{2}-4d^{2}\right)=4r^{2}-16d^{2} \] \[15d^{2}=3r^{2}, \qquad r^{2}=5d^{2}, \qquad r=d\sqrt{5}. \] So it happens for exactly one shape of picture: when the radius is \(\displaystyle \sqrt{5}\) times the distance of CD from the centre. For instance \(\displaystyle r=2\sqrt{5}\approx 4.47\ \mathrm{cm}\) with \(\displaystyle d=2\ \mathrm{cm}\) gives \[\mathrm{CD}=2\sqrt{20-4}=8\ \mathrm{cm}, \qquad \mathrm{AB}=2\sqrt{20-16}=4\ \mathrm{cm}, \] and indeed \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\). (Check \(\displaystyle 2d=4<4.47=r\), so AB really is a chord.) Every other choice of \(\displaystyle r\) and \(\displaystyle d\) gives a different ratio — and in fact, by pushing \(\displaystyle 2d\) close to \(\displaystyle r\), the ratio \(\displaystyle \mathrm{CD}/\mathrm{AB}\) can be made as large as you like.Answer: No, we cannot conclude \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\). Doubling the distance from the centre does not halve the chord. All that follows in general is \(\displaystyle \mathrm{CD}>\mathrm{AB}\); the equality \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) holds only in the special case \(\displaystyle r=d\sqrt{5}\).

The triangle formed by the chord and the centre turns out to be equilateral.Join OA and OB to get \(\displaystyle \triangle \mathrm{OAB}\), the triangle formed by the chord AB and the centre.Step $\displaystyle 1$ — it is isosceles. A and B lie on the circle, so \[\mathrm{OA}=\mathrm{OB}=12\ \mathrm{cm} \] (both are radii). By Exercise Set $\displaystyle 5.2$, Question $\displaystyle 1$, \(\displaystyle \triangle \mathrm{OAB}\) is isosceles with base AB, so its base angles are equal: \[\angle \mathrm{OAB}=\angle \mathrm{OBA}. \]Step $\displaystyle 2$ — find those base angles. Call each of them \(\displaystyle y\). The angles of a triangle add to \(\displaystyle 180^\circ\), and the angle at O is the given central angle \(\displaystyle 60^\circ\): \[60^\circ+y+y=180^\circ \] \[2y=120^\circ \] \[y=60^\circ. \]Step $\displaystyle 3$ — read off the chord. All three angles are now \(\displaystyle 60^\circ\), so \(\displaystyle \triangle \mathrm{OAB}\) is equiangular, and an equiangular triangle is equilateral. Hence all three sides are equal: \[\mathrm{AB}=\mathrm{OA}=\mathrm{OB}=12\ \mathrm{cm}. \]Sanity check with the chord formula. Drop the perpendicular from O to AB, meeting it at M. It bisects both the chord (Theorem $\displaystyle 5$) and the apex angle, so \(\displaystyle \angle \mathrm{AOM}=30^\circ\) and \(\displaystyle \triangle \mathrm{AOM}\) is a \(\displaystyle 30^\circ\)–\(\displaystyle 60^\circ\)–\(\displaystyle 90^\circ\) triangle. In such a triangle the side opposite \(\displaystyle 30^\circ\) is half the hypotenuse, so \(\displaystyle \mathrm{AM}=\tfrac{1}{2}\times 12=6\ \mathrm{cm}\) and \(\displaystyle \mathrm{AB}=12\ \mathrm{cm}\). The two methods agree.\(\displaystyle \mathrm{AB}=12\ \mathrm{cm}\) — the chord is as long as the radius.

All three parts are about Theorem $\displaystyle 9$ and how far its converse can be pushed.Throughout, X and Y are different from A and B, so that the angles \(\displaystyle \angle \mathrm{AXB}\) and \(\displaystyle \angle \mathrm{AYB}\) make sense. The chord AB cuts the circle into two arcs; for points on the circle, "on the same side of AB" and "on the same arc" mean the same thing.(i) No — there are no such points.Suppose X and Y are both on the circle and on the same side of AB, i.e. both on the same arc. Call the other arc, the one neither of them is on, arc AB. Then X and Y are both points of the circle lying outside arc AB, so Theorem $\displaystyle 9$ applies to each of them: \[\angle \mathrm{AXB}=\tfrac{1}{2}\left(\text{angle arc AB subtends at the centre}\right), \qquad \angle \mathrm{AYB}=\tfrac{1}{2}\left(\text{the same angle}\right). \] Both are half of one and the same central angle, so they are equal. That is the whole point of Theorem $\displaystyle 9$: the angle does not depend on which point of that arc you pick (Fig. $\displaystyle 5.23$). So no such pair X, Y exists.(ii) Not always — it fails exactly when AB is a diameter.Suppose X and Y are on the circle but on opposite arcs. Let the arc containing Y subtend the angle \(\displaystyle \alpha\) at the centre; then the arc containing X subtends the rest of a full turn, \(\displaystyle 360^\circ-\alpha\). Apply Theorem $\displaystyle 9$ to each point (each is outside the arc containing the other): \[\angle \mathrm{AXB}=\tfrac{\alpha}{2}, \qquad \angle \mathrm{AYB}=\tfrac{360^\circ-\alpha}{2}=180^\circ-\tfrac{\alpha}{2}. \] Adding, \[\angle \mathrm{AXB}+\angle \mathrm{AYB}=180^\circ. \] So points on opposite arcs always see AB at supplementary angles. Two supplementary angles are equal only when each is \(\displaystyle 90^\circ\), which needs \(\displaystyle \alpha=180^\circ\) — that is, AB is a diameter.Therefore:

If AB is not a diameter	\(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}\) does force X, Y onto the same arc
If AB is a diameter	every point of the circle sees AB at \(\displaystyle 90^\circ\)

The second row is the corollary to Theorem $\displaystyle 9$: the angle subtended by a diameter at any point of the circle is \(\displaystyle 90^\circ\). So if AB is a diameter, take X above AB and Y below it — then \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}=90^\circ\) although X and Y are on opposite sides. The statement as written is therefore not true in general.(iii) Yes if X and Y are on the same side of the line AB; not otherwise.Same side. A, B, X are not collinear, so by Theorem $\displaystyle 1$ there is exactly one circle through A, B and X; draw it. Suppose Y were not on it. Y is on the same side of AB as X, so Y is either inside or outside this new circle. Both alternatives are ruled out, so Y lies on the circle through A, B, X. This is exactly Theorem $\displaystyle 10$, which the book proves in the very next section, Section 5.8.Opposite sides. Then the circle through A, B, X need not pass through Y. For instance, suppose \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}=60^\circ\), with X above the line AB and Y below it. Draw the circle through A, B, X. Since \(\displaystyle \angle \mathrm{AXB}=60^\circ\), Theorem $\displaystyle 9$ says the lower arc AB (the one not containing X) subtends \(\displaystyle 2\times 60^\circ=120^\circ\) at the centre, so the upper arc subtends \(\displaystyle 360^\circ-120^\circ=240^\circ\). Applying Theorem $\displaystyle 9$ the other way round, a point of the lower arc sees AB at \(\displaystyle \tfrac{1}{2}\times 240^\circ=120^\circ\). Y is below AB but sees AB at only \(\displaystyle 60^\circ\), so Y is not one of those points: it is not on the circle. (Part (ii) is the same phenomenon — the two sides give supplementary angles, and \(\displaystyle 60^\circ\) and \(\displaystyle 120^\circ\) are the supplementary pair here.)Summary: (i) No, by Theorem $\displaystyle 9$ all points of one arc give the same angle. (ii) No, not always — it fails precisely when AB is a diameter, since then every point of the circle sees AB at \(\displaystyle 90^\circ\). (iii) Yes, provided X and Y are on the same side of AB (that is Theorem $\displaystyle 10$); if they are on opposite sides, the circle through A, B, X need not pass through Y.

Two arcs make up the whole circle, so their central angles add to \(\displaystyle 360^\circ\).Fig. $\displaystyle 5.26$ shows four points on a circle. Going round the circle they occur in the order A, D, C, B, and the chords AD, DC, CB, BA are drawn, making the $\displaystyle 4$-gon ADCB. The angle marked \(\displaystyle 100^\circ\) is \(\displaystyle \angle \mathrm{ADC}\), at the vertex D, and \(\displaystyle x\) is \(\displaystyle \angle \mathrm{ABC}\), at the vertex B. D and B are the two opposite vertices of this $\displaystyle 4$-gon.Let O be the centre of the circle. The chord AC splits the circle into two arcs: one goes from A to C through B, the other from A to C through D.Step $\displaystyle 1$ — use the \(\displaystyle 100^\circ\) angle to find a central angle. The point D lies on the circle, outside arc A–B–C. By Theorem $\displaystyle 9$, the angle that arc A–B–C subtends at the centre is double the angle it subtends at D: \[\text{central angle of arc A–B–C}=2\times \angle \mathrm{ADC}=2\times 100^\circ=200^\circ. \]Step $\displaystyle 2$ — the rest of the turn. Sweeping right round the centre is a full turn, \(\displaystyle 360^\circ\), and the two arcs together make up the whole circle. So \[\text{central angle of arc A–D–C}=360^\circ-200^\circ=160^\circ. \]Step $\displaystyle 3$ — come back out to B. The point B lies on the circle, outside arc A–D–C. By Theorem $\displaystyle 9$ again, the angle at B is half that arc's central angle: \[x=\angle \mathrm{ABC}=\tfrac{1}{2}\times 160^\circ=80^\circ. \]Check. \(\displaystyle 100^\circ+80^\circ=180^\circ\). That is no accident: ADCB is a cyclic $\displaystyle 4$-gon and D, B are opposite vertices, so Theorem $\displaystyle 11$ (in the very next section) says their angles must add to \(\displaystyle 180^\circ\). Once you have Theorem $\displaystyle 11$ the whole question is one line: \[x=180^\circ-100^\circ=80^\circ. \]\(\displaystyle x=80^\circ\).

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre and \(\displaystyle AB \) the chord. "Distance from the centre" always means the perpendicular distance, so drop \(\displaystyle OM \perp AB \), with \(\displaystyle OM = 5 \) cm and \(\displaystyle OA = 13 \) cm (a radius).The reason this helps is a theorem from this chapter: the perpendicular from the centre to a chord bisects the chord. So \(\displaystyle M \) is the midpoint of \(\displaystyle AB \), and \[AB = 2\,AM. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), so Pythagoras' theorem gives \[AM^{2} = OA^{2} - OM^{2} = 13^{2} - 5^{2} = 169 - 25 = 144, \] \[AM = 12 \text{ cm}. \]Therefore \(\displaystyle AB = 2 \times 12 = 24 \) cm.Check: \(\displaystyle 5,\,12,\,13 \) is a right triangle, since \(\displaystyle 25 + 144 = 169 \). And $\displaystyle 24$ cm is less than the diameter $\displaystyle 26$ cm, as any chord must be.The chord is $\displaystyle 24$ cm long.

Angle at the centre is twice the angle at the circle.Let the arc be \(\displaystyle AB \) and the centre \(\displaystyle O \), so that \(\displaystyle \angle AOB = 70^{\circ} \). Take a point \(\displaystyle P \) on the remaining part of the circle (that is, on the other arc, not on arc \(\displaystyle AB \) itself) and join \(\displaystyle PA \) and \(\displaystyle PB \).The theorem of this chapter says: the angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle. So \[\angle APB = \tfrac{1}{2}\,\angle AOB = \tfrac{1}{2}\times 70^{\circ} = 35^{\circ}. \]Two things worth noticing: The arc subtends \(\displaystyle 35^{\circ} \) at a point on the remaining part of the circle.

Perpendicular from the centre to a chord.The diameter is $\displaystyle 26$ cm, so the radius is \[r = \tfrac{26}{2} = 13 \text{ cm}. \]Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord of length $\displaystyle 24$ cm, and \(\displaystyle OM \perp AB \). Since the perpendicular from the centre bisects the chord, \[AM = \tfrac{24}{2} = 12 \text{ cm}. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), so \[OM^{2} = OA^{2} - AM^{2} = 13^{2} - 12^{2} = 169 - 144 = 25, \] \[OM = 5 \text{ cm}. \]Check by working backwards: a chord $\displaystyle 5$ cm from the centre of a $\displaystyle 13$ cm circle has half-length \(\displaystyle \sqrt{169-25} = 12 \), so full length $\displaystyle 24$ cm. Correct.The chord is $\displaystyle 5$ cm from the centre.

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord, and \(\displaystyle OM \perp AB \), so \(\displaystyle OM = 9 \) cm and \(\displaystyle OA = 15 \) cm.The perpendicular from the centre bisects the chord, so \(\displaystyle M \) is the midpoint of \(\displaystyle AB \) and triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \): \[AM^{2} = OA^{2} - OM^{2} = 15^{2} - 9^{2} = 225 - 81 = 144, \] \[AM = 12 \text{ cm}. \]Hence \[AB = 2 \times 12 = 24 \text{ cm}. \]Check: \(\displaystyle 9, 12, 15 \) is just the \(\displaystyle 3,4,5 \) triangle scaled by $\displaystyle 3$, and \(\displaystyle 81 + 144 = 225 \). Also \(\displaystyle 24 < 30 \), the diameter, as it must be.The chord is $\displaystyle 24$ cm long.

Points equidistant from two points lie on their perpendicular bisector.Let \(\displaystyle AB \) be any chord of a circle with centre \(\displaystyle O \).Short proof. \(\displaystyle OA \) and \(\displaystyle OB \) are both radii, so \[OA = OB. \] So \(\displaystyle O \) is a point equidistant from \(\displaystyle A \) and \(\displaystyle B \). But the set of all points equidistant from \(\displaystyle A \) and \(\displaystyle B \) is exactly the perpendicular bisector of \(\displaystyle AB \). Hence \(\displaystyle O \) lies on the perpendicular bisector of \(\displaystyle AB \) — that is, the perpendicular bisector of the chord passes through the centre.The same thing with congruent triangles, in case you would rather see every step. Let \(\displaystyle M \) be the midpoint of \(\displaystyle AB \), and join \(\displaystyle OA \), \(\displaystyle OB \), \(\displaystyle OM \). In triangles \(\displaystyle OMA \) and \(\displaystyle OMB \): So \(\displaystyle \triangle OMA \cong \triangle OMB \) by SSS, and therefore \(\displaystyle \angle OMA = \angle OMB \). These two angles form a linear pair, so they add up to \(\displaystyle 180^{\circ} \): \[\angle OMA + \angle OMB = 180^{\circ} \quad\Longrightarrow\quad 2\,\angle OMA = 180^{\circ} \quad\Longrightarrow\quad \angle OMA = 90^{\circ}. \]So the line \(\displaystyle OM \) passes through the midpoint of \(\displaystyle AB \) and is perpendicular to \(\displaystyle AB \) — it is the perpendicular bisector of \(\displaystyle AB \). Since a segment has only one perpendicular bisector, the centre \(\displaystyle O \) lies on it.Why this is useful: if you are handed a circle with no marked centre, draw any two non-parallel chords and construct their perpendicular bisectors. Each must pass through the centre, so the point where they meet is the centre.Hence the perpendicular bisector of a chord always passes through the centre of the circle.

Angle in a semicircle.Let \(\displaystyle O \) be the centre. Since \(\displaystyle AB \) is a diameter, \(\displaystyle A \), \(\displaystyle O \), \(\displaystyle B \) lie on one straight line, so \[\angle AOB = 180^{\circ}. \]Reason $\displaystyle 1$ (using the central-angle theorem). The arc \(\displaystyle AB \) that does not contain \(\displaystyle C \) is a semicircle, and it subtends \(\displaystyle 180^{\circ} \) at the centre. An arc subtends at a point on the remaining part of the circle half the angle it subtends at the centre, so \[\angle ACB = \tfrac{1}{2} \times 180^{\circ} = 90^{\circ}. \]Reason $\displaystyle 2$ (from scratch, using only isosceles triangles). Join \(\displaystyle OC \). Then \(\displaystyle OA = OB = OC \), all radii. Now in triangle \(\displaystyle ABC \), the three angles are \(\displaystyle \angle A = x \), \(\displaystyle \angle B = y \), and \(\displaystyle \angle ACB = \angle OCA + \angle OCB = x + y \). Their sum is \(\displaystyle 180^{\circ} \): \[x + y + (x+y) = 180^{\circ} \quad\Longrightarrow\quad 2(x+y) = 180^{\circ} \quad\Longrightarrow\quad x + y = 90^{\circ}. \] So \(\displaystyle \angle ACB = 90^{\circ} \).This is true wherever \(\displaystyle C \) sits on the circle (as long as \(\displaystyle C \) is not \(\displaystyle A \) or \(\displaystyle B \)) — sliding \(\displaystyle C \) around changes \(\displaystyle x \) and \(\displaystyle y \), but never their sum.\(\displaystyle \angle ACB = 90^{\circ} \): the angle in a semicircle is a right angle.

Opposite angles of a cyclic quadrilateral are supplementary.Why the rule is true. \(\displaystyle \angle A \) stands on the arc \(\displaystyle BCD \) (the arc from \(\displaystyle B \) to \(\displaystyle D \) not containing \(\displaystyle A \)), and \(\displaystyle \angle C \) stands on the arc \(\displaystyle DAB \). Each inscribed angle is half its arc, and the two arcs together make the whole circle, \(\displaystyle 360^{\circ} \). So \[\angle A + \angle C = \tfrac{1}{2}\big(\text{arc } BCD + \text{arc } DAB\big) = \tfrac{1}{2}\times 360^{\circ} = 180^{\circ}. \] The same argument gives \(\displaystyle \angle B + \angle D = 180^{\circ} \).Applying it. \[\angle C = 180^{\circ} - \angle A = 180^{\circ} - 75^{\circ} = 105^{\circ}, \] \[\angle D = 180^{\circ} - \angle B = 180^{\circ} - 110^{\circ} = 70^{\circ}. \]Check: the four angles of any quadrilateral add to \(\displaystyle 360^{\circ} \), and indeed \[75^{\circ} + 110^{\circ} + 105^{\circ} + 70^{\circ} = 360^{\circ}. \]\(\displaystyle \angle C = 105^{\circ} \) and \(\displaystyle \angle D = 70^{\circ} \).

Opposite angles of a cyclic quadrilateral are supplementary.In the cyclic quadrilateral \(\displaystyle PQRS \), the angles \(\displaystyle \angle P \) and \(\displaystyle \angle R \) are opposite, so \[\angle P + \angle R = 180^{\circ}. \]Substituting the given expressions: \[(2x + 10)^{\circ} + (3x - 20)^{\circ} = 180^{\circ}, \] \[5x - 10 = 180, \] \[5x = 190, \] \[x = 38. \]Now put \(\displaystyle x = 38 \) back into each expression: \[\angle P = (2 \times 38 + 10)^{\circ} = (76 + 10)^{\circ} = 86^{\circ}, \] \[\angle R = (3 \times 38 - 20)^{\circ} = (114 - 20)^{\circ} = 94^{\circ}. \]Check: \(\displaystyle 86^{\circ} + 94^{\circ} = 180^{\circ} \). Both angles are positive and less than \(\displaystyle 180^{\circ} \), so this is a genuine quadrilateral.\(\displaystyle x = 38 \), \(\displaystyle \angle P = 86^{\circ} \), \(\displaystyle \angle R = 94^{\circ} \).

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord of length $\displaystyle 16$ cm, and \(\displaystyle OM \perp AB \), with \(\displaystyle OM = 6 \) cm.The perpendicular from the centre bisects the chord, so \[AM = \tfrac{16}{2} = 8 \text{ cm}. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), and \(\displaystyle OA \) is the radius \(\displaystyle r \): \[r^{2} = OM^{2} + AM^{2} = 6^{2} + 8^{2} = 36 + 64 = 100, \] \[r = 10 \text{ cm}. \]Check: \(\displaystyle 6, 8, 10 \) is the \(\displaystyle 3,4,5 \) triangle doubled. And the radius $\displaystyle 10$ cm must be more than half the chord ($\displaystyle 8$ cm) — it is.The radius of the circle is $\displaystyle 10$ cm.

Equal chords cut off equal arcs, then split the quadrilateral by a diagonal.Call the quadrilateral \(\displaystyle ABCD \), taken round the circle with \[AB = 5,\quad BC = 5,\quad CD = 12,\quad DA = 12. \]Step $\displaystyle 1$: the arcs. In one circle, equal chords cut off equal arcs. Since \(\displaystyle AB = BC \), arc \(\displaystyle AB = \) arc \(\displaystyle BC \); call each of them \(\displaystyle p \). Since \(\displaystyle CD = DA \), arc \(\displaystyle CD = \) arc \(\displaystyle DA \); call each of them \(\displaystyle q \). Going once round the circle uses up all four arcs: \[p + p + q + q = 360^{\circ} \quad\Longrightarrow\quad 2(p+q) = 360^{\circ} \quad\Longrightarrow\quad p + q = 180^{\circ}. \]Step $\displaystyle 2$: a right angle appears. The angle \(\displaystyle \angle DAB \) is the inscribed angle at \(\displaystyle A \). It stands on the arc going from \(\displaystyle B \) to \(\displaystyle D \) the long way round through \(\displaystyle C \), which is arc \(\displaystyle BC \) + arc \(\displaystyle CD \) \(\displaystyle = p + q = 180^{\circ} \). An inscribed angle is half its arc, so \[\angle DAB = \tfrac{1}{2}\times 180^{\circ} = 90^{\circ}. \] And since \(\displaystyle ABCD \) is cyclic, \(\displaystyle \angle BCD = 180^{\circ} - 90^{\circ} = 90^{\circ} \) as well.Step $\displaystyle 3$: two right triangles. Draw the diagonal \(\displaystyle BD \). It cuts \(\displaystyle ABCD \) into \(\displaystyle \triangle ABD \) and \(\displaystyle \triangle CBD \), and both have a right angle: (As a bonus, \(\displaystyle \angle DAB = 90^{\circ} \) means \(\displaystyle BD \) is a diameter, so the circle has radius \(\displaystyle 6.5 \) units.)Step $\displaystyle 4$: add the areas. \[\text{Area} = \tfrac{1}{2}(5)(12) + \tfrac{1}{2}(5)(12) = 30 + 30 = 60. \]What if the sides come in the order \(\displaystyle 5, 12, 5, 12 \) instead? Then opposite sides are equal, so the quadrilateral is a parallelogram; and a parallelogram inscribed in a circle must be a rectangle (see the question on that). A \(\displaystyle 5 \times 12 \) rectangle has area \(\displaystyle 5 \times 12 = 60 \) — the same answer. That is not a coincidence: for a cyclic quadrilateral the area depends only on the four side lengths, not on the order they are arranged in.The area is $\displaystyle 60$ square units.

Look at the arc that each side cuts off.The idea. Take one side, say \(\displaystyle AB \). The line \(\displaystyle AB \) splits the plane in two, and the other two vertices \(\displaystyle C \) and \(\displaystyle D \) are on one of those sides — the inside of the quadrilateral is on that side too. The centre \(\displaystyle O \) lies on the same side of the chord \(\displaystyle AB \) as the major arc. So:\[O \text{ is inside } ABCD \iff \text{every side cuts off an arc smaller than a semicircle.} \]Turning that into something you can measure. The arc \(\displaystyle AB \) that does not contain \(\displaystyle C \) and \(\displaystyle D \) is measured by the angle it subtends at \(\displaystyle C \) (or at \(\displaystyle D \)): \(\displaystyle \angle ACB = \tfrac{1}{2}\,\text{arc } AB \). Therefore, for each side, look at the angle it subtends at one of the two opposite vertices: So one honest method is: check all four such angles. All acute \(\displaystyle \Rightarrow \) centre inside.The best way — you only ever need one measurement. The four arcs add up to \(\displaystyle 360^{\circ} \), so at most one of them can exceed \(\displaystyle 180^{\circ} \). Moreover, the side cutting off that big arc is always the longest side. Here is why. Suppose arc \(\displaystyle AB = a > 180^{\circ} \). Then the other three arcs satisfy \(\displaystyle b + c + d = 360^{\circ} - a < 180^{\circ} \). Now the chord \(\displaystyle AB \) also spans the minor arc of measure \(\displaystyle 360^{\circ} - a = b+c+d \), and each of \(\displaystyle b, c, d \) is smaller than that. Among minor arcs (arcs under \(\displaystyle 180^{\circ} \)) of one circle, a longer arc has a longer chord. Hence \(\displaystyle AB \) is longer than \(\displaystyle BC \), \(\displaystyle CD \) and \(\displaystyle DA \).So the quick test is:1. Find the longest side of the quadrilateral. 2. Measure the angle it subtends at either of the two remaining vertices (the two answers are equal). 3. Acute \(\displaystyle \Rightarrow \) the centre is inside; exactly \(\displaystyle 90^{\circ} \Rightarrow \) the centre is on that side (its midpoint); obtuse \(\displaystyle \Rightarrow \) the centre is outside, beyond that side.Sanity checks. For a rectangle every side subtends an acute angle at the far vertices, and indeed the centre is inside (it is where the diagonals cross). For a thin quadrilateral whose four vertices are bunched together on a small piece of the circle, the long "closing" side subtends a fat obtuse angle, and the centre is far outside.An equivalent way of saying it, if you prefer triangles: a diagonal cuts the quadrilateral into two triangles inscribed in the same circle, and the circumcentre of a triangle lies inside it exactly when the triangle is acute-angled. The centre lies inside the quadrilateral precisely when at least one of those two triangles is not obtuse.Test the longest side: if it subtends an acute angle at the opposite vertices the centre is inside, if a right angle the centre lies on that side, and if an obtuse angle the centre is outside. Since at most one side can cut off a major arc, and that side must be the longest one, this single check settles it.

Equal chords are equidistant from the centre, then RHS congruence.Let the circle have centre \(\displaystyle O \) and radius \(\displaystyle r \). Let the two chords be \(\displaystyle AB \) and \(\displaystyle CD \) with \(\displaystyle AB = CD \), crossing at a point \(\displaystyle P \) inside the circle. Let \(\displaystyle M \) be the midpoint of \(\displaystyle AB \) and \(\displaystyle N \) the midpoint of \(\displaystyle CD \), and join \(\displaystyle OM \), \(\displaystyle ON \), \(\displaystyle OP \).Step $\displaystyle 1$: \(\displaystyle OM \perp AB \) and \(\displaystyle ON \perp CD \). The line from the centre to the midpoint of a chord is perpendicular to that chord (this is the perpendicular-bisector result of this chapter).Step $\displaystyle 2$: the two chords are the same distance from the centre. Write \(\displaystyle h \) for the common half-length, so \[h = AM = MB = CN = ND = \tfrac{1}{2}AB = \tfrac{1}{2}CD . \] In the right triangles \(\displaystyle OMA \) and \(\displaystyle ONC \), Pythagoras gives \[OM^{2} = r^{2} - h^{2} = ON^{2}, \qquad\text{so}\qquad OM = ON. \]Step $\displaystyle 3$: \(\displaystyle P \) is the same distance from both midpoints. Compare \(\displaystyle \triangle OMP \) and \(\displaystyle \triangle ONP \): By RHS congruence, \(\displaystyle \triangle OMP \cong \triangle ONP \), so \[MP = NP. \] Call this common distance \(\displaystyle d \).Step $\displaystyle 4$: read off the four pieces. On the chord \(\displaystyle AB \), the point \(\displaystyle P \) sits at distance \(\displaystyle d \) from the midpoint \(\displaystyle M \), so the two pieces of \(\displaystyle AB \) are \[h + d \quad\text{and}\quad h - d. \] On the chord \(\displaystyle CD \), the point \(\displaystyle P \) sits at the same distance \(\displaystyle d \) from the midpoint \(\displaystyle N \), so the two pieces of \(\displaystyle CD \) are also \[h + d \quad\text{and}\quad h - d. \]So the pieces of one chord match the pieces of the other: one of \(\displaystyle PC \), \(\displaystyle PD \) equals \(\displaystyle PA \), and the other equals \(\displaystyle PB \). (Which name goes with which just depends on how you labelled the endpoints.)Special case: if \(\displaystyle d = 0 \) then \(\displaystyle P \) is the midpoint of both chords, so both chords are bisected at \(\displaystyle P \) — which happens exactly when both are diameters and \(\displaystyle P = O \).An alternative, if you have met the intersecting-chords relation \(\displaystyle PA \cdot PB = PC \cdot PD \): here also \(\displaystyle PA + PB = AB = CD = PC + PD \). Two positive numbers are completely determined by their sum and their product, so the pair \(\displaystyle \{PA, PB\} \) must be the same pair as \(\displaystyle \{PC, PD\} \).Hence if two intersecting chords are equal in length, the segments of one are equal to the corresponding segments of the other.

Build the right triangle first; its circumcircle is the circle you want.First, what circle are we after? Let \(\displaystyle AB \) be the required chord, \(\displaystyle AB = 6 \) cm, and let \(\displaystyle O \) be the centre with \(\displaystyle OM \perp AB \), \(\displaystyle OM = 3 \) cm. Then \(\displaystyle AM = 3 \) cm, so triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \) with both legs equal to $\displaystyle 3$ cm, and \[OA^{2} = 3^{2} + 3^{2} = 18, \qquad OA = 3\sqrt{2} \approx 4.2 \text{ cm}. \] Also \(\displaystyle \angle AOM = 45^{\circ} \), so \(\displaystyle \angle AOB = 90^{\circ} \): the chord subtends a right angle at the centre.The hint's construction. A chord subtending \(\displaystyle 90^{\circ} \) at the centre is a leg of an inscribed right isosceles triangle. So take the triangle with a right angle and two equal sides of $\displaystyle 6$ cm, and draw its circumcircle:1. Draw a right angle at a point \(\displaystyle A \). 2. On one arm mark \(\displaystyle B \) with \(\displaystyle AB = 6 \) cm; on the other arm mark \(\displaystyle D \) with \(\displaystyle AD = 6 \) cm. 3. Join \(\displaystyle BD \). (It is the hypotenuse, \(\displaystyle BD = \sqrt{6^{2}+6^{2}} = 6\sqrt{2} \approx 8.5 \) cm.) 4. Mark \(\displaystyle O \), the midpoint of \(\displaystyle BD \). Because the angle in a semicircle is a right angle, the hypotenuse of a right triangle is a diameter of its circumcircle, so the circumcentre is the midpoint of \(\displaystyle BD \). 5. With centre \(\displaystyle O \) and radius \(\displaystyle OA = OB = OD = 3\sqrt{2} \approx 4.2 \) cm, draw the circle. It passes through \(\displaystyle A \), \(\displaystyle B \) and \(\displaystyle D \).Then \(\displaystyle AB \) is a chord of this circle of length $\displaystyle 6$ cm, and its distance from \(\displaystyle O \) is $\displaystyle 3$ cm — exactly what was asked. (So is \(\displaystyle AD \).)Verification with coordinates. Put \(\displaystyle A = (0,0) \), \(\displaystyle B = (6,0) \), \(\displaystyle D = (0,6) \). Then \(\displaystyle O \), the midpoint of \(\displaystyle BD \), is \(\displaystyle (3,3) \). Distances: \(\displaystyle OA = OB = OD = \sqrt{3^{2}+3^{2}} = 3\sqrt{2} \), so all three points really are on one circle centred at \(\displaystyle O \). The chord \(\displaystyle AB \) lies along the \(\displaystyle x \)-axis, so its distance from \(\displaystyle O = (3,3) \) is \(\displaystyle 3 \) cm, and its length is \(\displaystyle 6 \) cm. Correct.A shorter route, if you do not want a triangle at all: draw \(\displaystyle AB = 6 \) cm, construct its perpendicular bisector, mark \(\displaystyle O \) on that bisector $\displaystyle 3$ cm from \(\displaystyle AB \), and draw the circle with centre \(\displaystyle O \) and radius \(\displaystyle OA \). This works because the centre of any circle must lie on the perpendicular bisector of each of its chords.Draw a right isosceles triangle with legs $\displaystyle 6$ cm and take its circumcircle: centre at the midpoint of the hypotenuse, radius \(\displaystyle 3\sqrt{2} \approx 4.2 \) cm. Each $\displaystyle 6$ cm leg is then a chord $\displaystyle 3$ cm from the centre.

Combine "opposite angles of a parallelogram are equal" with "opposite angles of a cyclic quadrilateral are supplementary".Suppose a parallelogram \(\displaystyle ABCD \) is inscribed in a circle. Two facts now apply to the same pair of angles.Because \(\displaystyle ABCD \) is a parallelogram, opposite angles are equal: \[\angle A = \angle C. \]Because \(\displaystyle ABCD \) is cyclic, opposite angles are supplementary: \[\angle A + \angle C = 180^{\circ}. \]Substituting the first into the second, \[\angle A + \angle A = 180^{\circ} \quad\Longrightarrow\quad 2\angle A = 180^{\circ} \quad\Longrightarrow\quad \angle A = 90^{\circ}, \] and hence \(\displaystyle \angle C = 90^{\circ} \) too. The same argument on the other pair gives \(\displaystyle \angle B = \angle D = 90^{\circ} \). (Or simply: the angles of a quadrilateral add to \(\displaystyle 360^{\circ} \), so the remaining two share \(\displaystyle 180^{\circ} \) equally.)A parallelogram with all its angles right angles is by definition a rectangle. So the only parallelogram that can be inscribed in a circle is a rectangle.And a rectangle really can be. This matters — otherwise the statement would be empty. Let \(\displaystyle ABCD \) be a rectangle and \(\displaystyle P \) the point where its diagonals meet. The diagonals of a rectangle are equal and bisect each other, so \[PA = PB = PC = PD = \tfrac{1}{2}(\text{diagonal}). \] All four vertices are the same distance from \(\displaystyle P \), so the circle with centre \(\displaystyle P \) and that radius passes through all four — the rectangle is cyclic.Every cyclic parallelogram is a rectangle, and every rectangle is cyclic; so the rectangle is the only parallelogram that can be inscribed in a circle.

A right angle inscribed in a circle stands on a diameter.Let the rectangle \(\displaystyle ABCD \) be inscribed in a circle with centre \(\displaystyle O \), and let \(\displaystyle P \) be the point where its diagonals \(\displaystyle AC \) and \(\displaystyle BD \) cross. We must show \(\displaystyle P = O \).Step $\displaystyle 1$: each diagonal is a diameter. Look at \(\displaystyle \angle ABC \). It is an angle of a rectangle, so \(\displaystyle \angle ABC = 90^{\circ} \); and it is an inscribed angle standing on the arc \(\displaystyle AC \) not containing \(\displaystyle B \). An inscribed angle is half its arc, so \[\text{arc } AC = 2 \times 90^{\circ} = 180^{\circ}. \] An arc of \(\displaystyle 180^{\circ} \) is a semicircle, so its chord \(\displaystyle AC \) is a diameter, and therefore \(\displaystyle AC \) passes through \(\displaystyle O \).The identical argument with \(\displaystyle \angle BCD = 90^{\circ} \) shows that \(\displaystyle BD \) is a diameter too, so \(\displaystyle BD \) also passes through \(\displaystyle O \).Step $\displaystyle 2$: finish. Both diagonals pass through \(\displaystyle O \). Two distinct lines meet in at most one point, and the diagonals of a rectangle are distinct lines meeting at \(\displaystyle P \). Hence \[P = O. \]A second proof, using distances. In a rectangle the diagonals are equal and bisect each other, so \[PA = PB = PC = PD = \tfrac{1}{2}(\text{diagonal}). \] So \(\displaystyle P \) is equidistant from all four vertices, i.e. \(\displaystyle P \) is the centre of a circle through \(\displaystyle A, B, C, D \). But three non-collinear points lie on exactly one circle, so that circle is the given circle and \(\displaystyle P \) is its centre.The diagonals of an inscribed rectangle are diameters, so their point of intersection is the centre of the circle.

Every such midpoint is the same distance from the centre.Let the circle have centre \(\displaystyle O \) and radius \(\displaystyle r \), and let all the chords have the same fixed length \(\displaystyle \ell \) (with \(\displaystyle \ell \le 2r \), since no chord can be longer than a diameter).One direction: every midpoint is at a fixed distance from \(\displaystyle O \). Take any chord \(\displaystyle AB \) of length \(\displaystyle \ell \), with midpoint \(\displaystyle M \). The line from the centre to the midpoint of a chord is perpendicular to the chord, so \(\displaystyle \triangle OMA \) is right angled at \(\displaystyle M \), with \(\displaystyle AM = \tfrac{\ell}{2} \) and \(\displaystyle OA = r \). Pythagoras gives \[OM^{2} = r^{2} - \left(\frac{\ell}{2}\right)^{2}, \qquad OM = \sqrt{\,r^{2} - \frac{\ell^{2}}{4}\,}. \] Nothing on the right depends on which chord of length \(\displaystyle \ell \) we chose. So every one of these midpoints lies at this one fixed distance \(\displaystyle d = \sqrt{r^{2} - \ell^{2}/4} \) from \(\displaystyle O \) — that is, on a circle with centre \(\displaystyle O \) and radius \(\displaystyle d \).The other direction: the whole of that circle is used. Take any point \(\displaystyle M \) with \(\displaystyle OM = d \). Draw the line through \(\displaystyle M \) perpendicular to \(\displaystyle OM \); it cuts the circle at two points \(\displaystyle A \) and \(\displaystyle B \). Since \(\displaystyle OM \perp AB \), \(\displaystyle M \) is the midpoint of \(\displaystyle AB \), and \[AM^{2} = r^{2} - d^{2} = r^{2} - \left(r^{2} - \frac{\ell^{2}}{4}\right) = \frac{\ell^{2}}{4}, \] so \(\displaystyle AB = 2\,AM = \ell \). So \(\displaystyle M \) really is the midpoint of a chord of length \(\displaystyle \ell \). Nothing on that circle is left out.Example: in a circle of radius $\displaystyle 10$ units, the midpoints of all chords of length $\displaystyle 12$ units form a circle of radius \(\displaystyle \sqrt{100-36} = 8 \) units around the same centre.The degenerate case: if \(\displaystyle \ell = 2r \) the chords are diameters, they all have the same midpoint \(\displaystyle O \), and the "circle" of midpoints shrinks to the single point \(\displaystyle O \) (radius \(\displaystyle 0 \)) — which the formula gives correctly.The midpoints form a circle with the same centre as the given circle (a concentric circle), of radius \(\displaystyle \sqrt{\,r^{2} - \ell^{2}/4\,} \).

SSS congruence on the two triangles the radii make.We are given a circle with centre \(\displaystyle O \) and two chords from the same point \(\displaystyle A \), with \(\displaystyle AB = AC \). Join \(\displaystyle OA \), \(\displaystyle OB \), \(\displaystyle OC \).Compare \(\displaystyle \triangle OAB \) and \(\displaystyle \triangle OAC \): So \(\displaystyle \triangle OAB \cong \triangle OAC \) by SSS. Corresponding angles of congruent triangles are equal, so \[\angle OAB = \angle OAC. \]But \(\displaystyle \angle OAB \) and \(\displaystyle \angle OAC \) are the two parts into which the ray \(\displaystyle AO \) divides \(\displaystyle \angle BAC \). Since these two parts are equal, \(\displaystyle AO \) is the bisector of \(\displaystyle \angle BAC \). Hence \(\displaystyle O \) lies on that bisector, which is the statement we wanted.Another way to see it. Both \(\displaystyle A \) and \(\displaystyle O \) are equidistant from \(\displaystyle B \) and \(\displaystyle C \) — \(\displaystyle A \) because \(\displaystyle AB = AC \), and \(\displaystyle O \) because \(\displaystyle OB = OC \) are radii. Two distinct points equidistant from \(\displaystyle B \) and \(\displaystyle C \) determine the perpendicular bisector of \(\displaystyle BC \), so the line \(\displaystyle AO \) is the perpendicular bisector of \(\displaystyle BC \). In the isosceles triangle \(\displaystyle ABC \) (with \(\displaystyle AB = AC \)), the perpendicular bisector of the base is also the bisector of the apex angle \(\displaystyle \angle BAC \). So \(\displaystyle O \) lies on the bisector of \(\displaystyle \angle BAC \).A caution worth noticing: this says the centre is on the bisector, not that it is inside the triangle. If \(\displaystyle \angle BAC \) is obtuse the centre still lies on the bisecting ray, just beyond \(\displaystyle BC \).Because \(\displaystyle OB = OC \) and \(\displaystyle AB = AC \), the triangles \(\displaystyle OAB \) and \(\displaystyle OAC \) are congruent, so \(\displaystyle AO \) splits \(\displaystyle \angle BAC \) into two equal parts — the centre lies on the angle bisector.

Two right triangles sharing the same radius.Let \(\displaystyle O \) be the centre and \(\displaystyle r \) the radius. Let \(\displaystyle AB = 24 \) cm and \(\displaystyle CD = 10 \) cm be the two parallel chords, both on the same side of \(\displaystyle O \). Drop the perpendicular from \(\displaystyle O \); because the chords are parallel, one single perpendicular line meets both, at \(\displaystyle M \) on \(\displaystyle AB \) and \(\displaystyle N \) on \(\displaystyle CD \), and these are the midpoints of the chords.Write \(\displaystyle OM = p \) and \(\displaystyle ON = q \). Half-lengths are \(\displaystyle AM = 12 \) cm and \(\displaystyle CN = 5 \) cm, so Pythagoras in the two right triangles \(\displaystyle OMA \) and \(\displaystyle ONC \) gives \[r^{2} = p^{2} + 12^{2} = p^{2} + 144, \tag{1} \] \[r^{2} = q^{2} + 5^{2} = q^{2} + 25. \tag{2} \]Setting up the $\displaystyle 7$ cm. Both chords are on the same side of \(\displaystyle O \), so \(\displaystyle M \) and \(\displaystyle N \) are on the same side and the gap between the chords is the difference of the distances: \[|q - p| = 7. \] The longer chord is the nearer one to the centre, so \(\displaystyle p < q \) and \[q = p + 7. \]Solving. Put \(\displaystyle q = p+7 \) into ($\displaystyle 2$) and set it equal to ($\displaystyle 1$): \[p^{2} + 144 = (p+7)^{2} + 25 = p^{2} + 14p + 49 + 25, \] \[144 = 14p + 74, \] \[14p = 70, \qquad p = 5. \] Then \(\displaystyle q = 5 + 7 = 12 \), and from ($\displaystyle 1$) \[r^{2} = 5^{2} + 144 = 25 + 144 = 169, \qquad r = 13 \text{ cm}. \]Check the other possibility, to be safe. If instead we had assumed the $\displaystyle 24$ cm chord was the farther one, \(\displaystyle p = q + 7 \), then \(\displaystyle (q+7)^{2} + 144 = q^{2} + 25 \) gives \(\displaystyle 14q + 49 + 144 = 25 \), i.e. \(\displaystyle 14q = -168 \), a negative distance. Impossible — so the solution above is the only one.Check the answer directly. With \(\displaystyle r = 13 \): the chord $\displaystyle 5$ cm from the centre has half-length \(\displaystyle \sqrt{169-25} = 12 \), so length $\displaystyle 24$ cm; the chord $\displaystyle 12$ cm from the centre has half-length \(\displaystyle \sqrt{169-144}=5 \), so length $\displaystyle 10$ cm; and the two are \(\displaystyle 12 - 5 = 7 \) cm apart on the same side. All three conditions hold.The radius of the circle is $\displaystyle 13$ cm.

Cut the hexagon into six triangles at the centre.Let the circle have centre \(\displaystyle O\) and radius \(\displaystyle r\), and let the regular hexagon be \(\displaystyle ABCDEF\) with all six vertices on the circle. Join \(\displaystyle O\) to every vertex. This cuts the hexagon into $\displaystyle 6$ triangles, and every one of the joining segments is a radius: \[OA = OB = OC = OD = OE = OF = r. \]Step $\displaystyle 1$: each angle at the centre is \(\displaystyle 60^\circ\). The hexagon is regular, so its six sides are equal; together with the equal radii, the six triangles are congruent by SSS. Hence the six angles they make at \(\displaystyle O\) are equal. Those six angles fill up one complete turn about \(\displaystyle O\), so \[\text{each angle at } O = \frac{360^\circ}{6} = 60^\circ. \]Step $\displaystyle 2$: each triangle is equilateral, so the side equals \(\displaystyle r\). Take \(\displaystyle \triangle OAB\). Since \(\displaystyle OA = OB\), it is isosceles, so the two base angles are equal: \[\angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ. \] All three angles are \(\displaystyle 60^\circ\), so \(\displaystyle \triangle OAB\) is equilateral and \[AB = OA = r. \]Step $\displaystyle 3$: the distance of a side from the centre. Drop \(\displaystyle OM \perp AB\), with \(\displaystyle M\) on \(\displaystyle AB\). The perpendicular from the centre to a chord bisects the chord (the same thing happens here because \(\displaystyle \triangle OAB\) is isosceles), so \[AM = \frac{AB}{2} = \frac{r}{2}. \] Now use Pythagoras in the right triangle \(\displaystyle OMA\): \[OM^2 = OA^2 - AM^2 = r^2 - \frac{r^2}{4} = \frac{3r^2}{4}, \qquad OM = \frac{\sqrt{3}}{2}\,r. \]A quick sanity check. The perimeter of the hexagon is \(\displaystyle 6r\), and the circumference of the circle is \(\displaystyle 2\pi r \approx 6.28r\). The hexagon sits just inside the circle, so its perimeter should be a little less than the circumference — and \(\displaystyle 6r < 6.28r\). Also \(\displaystyle OM = 0.866r\) is less than \(\displaystyle r\), as it must be, since a chord lies nearer the centre than the circle does.Each side of the hexagon has length \(\displaystyle r\), and each side is at a distance \(\displaystyle \dfrac{\sqrt{3}}{2}r \approx 0.87r\) from the centre.

Angles in the same segment.(Note that here \(\displaystyle O\) is one of the four vertices of the quadrilateral, not the centre of the circle.)The four points lie on the circle in the order \(\displaystyle M, N, O, P\). Draw the diagonal \(\displaystyle MP\). Now look at the two angles asked about: \(\displaystyle \angle \mathrm{MNP}\) is the angle that the same* chord \(\displaystyle MP\) subtends at the point \(\displaystyle N\).The chord \(\displaystyle MP\) splits the circle into two arcs. Going round the circle in order \(\displaystyle M \to N \to O \to P\), both \(\displaystyle N\) and \(\displaystyle O\) lie on the same one of these two arcs — that is, \(\displaystyle N\) and \(\displaystyle O\) are points of the same segment standing on \(\displaystyle MP\).Angles in the same segment are equal, because each of them is half of the one central angle standing on the arc \(\displaystyle MP\). So \[\angle \mathrm{MOP} = \angle \mathrm{MNP}. \]Where the diameter comes in. The equality above did not actually need \(\displaystyle MN\) to be a diameter. What the diameter tells you is the size of these angles. Since \(\displaystyle MN\) is a diameter, the angle in the semicircle at \(\displaystyle P\) is a right angle: \[\angle \mathrm{MPN} = 90^\circ . \] So in the right triangle \(\displaystyle MPN\), \[\angle \mathrm{MNP} = 90^\circ - \angle \mathrm{PMN}, \] and therefore \[\angle \mathrm{MOP} = \angle \mathrm{MNP} = 90^\circ - \angle \mathrm{PMN}. \] (For the same reason \(\displaystyle \angle \mathrm{MON} = 90^\circ\) as well, since \(\displaystyle O\) also sees the diameter \(\displaystyle MN\).)Checked numerically: placing \(\displaystyle M\) and \(\displaystyle N\) at the two ends of a diameter and trying many positions of \(\displaystyle O\) and \(\displaystyle P\) on the arc, the three quantities \(\displaystyle \angle \mathrm{MOP}\), \(\displaystyle \angle \mathrm{MNP}\) and \(\displaystyle 90^\circ - \angle \mathrm{PMN}\) agreed every time (for example \(\displaystyle 13.47^\circ\), \(\displaystyle 43.62^\circ\), \(\displaystyle 81.98^\circ\) in three different positions).\(\displaystyle \angle \mathrm{MOP} = \angle \mathrm{MNP}\): they are equal, being angles in the same segment standing on the chord \(\displaystyle MP\); and because \(\displaystyle MN\) is a diameter each of them equals \(\displaystyle 90^\circ - \angle \mathrm{PMN}\).

A linear pair, plus the fact that opposite angles of a cyclic quadrilateral add to \(\displaystyle 180^\circ\).First, a word about the labelling in the example. For \(\displaystyle \angle \mathrm{CDE}\) to be an exterior angle at the vertex \(\displaystyle D\), the point \(\displaystyle E\) must lie on \(\displaystyle AD\) produced beyond \(\displaystyle D\) — that is, you extend the side \(\displaystyle AD\). (If \(\displaystyle E\) were taken on \(\displaystyle CD\) produced, then \(\displaystyle C, D, E\) would be collinear and \(\displaystyle \angle \mathrm{CDE}\) would just be a straight angle of \(\displaystyle 180^\circ\); in that picture the exterior angle at \(\displaystyle D\) is \(\displaystyle \angle \mathrm{ADE}\).) Both choices give the same result, as the last step below shows.Step $\displaystyle 1$ — opposite angles of a cyclic quadrilateral. For the cyclic quadrilateral \(\displaystyle ABCD\), \[\angle \mathrm{ABC} + \angle \mathrm{ADC} = 180^\circ . \] (Reason: \(\displaystyle \angle \mathrm{ABC}\) is half the central angle standing on the arc \(\displaystyle ADC\), and \(\displaystyle \angle \mathrm{ADC}\) is half the central angle standing on the arc \(\displaystyle ABC\). Those two central angles together make one full turn, \(\displaystyle 360^\circ\), so the two inscribed angles add to \(\displaystyle \tfrac{1}{2}\times 360^\circ = 180^\circ\).)Step $\displaystyle 2$ — a linear pair at \(\displaystyle D\). Extend \(\displaystyle AD\) beyond \(\displaystyle D\) to the point \(\displaystyle E\). Then \(\displaystyle A, D, E\) lie on a straight line, so the angles \(\displaystyle \angle \mathrm{ADC}\) and \(\displaystyle \angle \mathrm{CDE}\) sit on that straight line at \(\displaystyle D\) and form a linear pair: \[\angle \mathrm{ADC} + \angle \mathrm{CDE} = 180^\circ . \]Step $\displaystyle 3$ — compare. Both sums equal \(\displaystyle 180^\circ\): \[\angle \mathrm{ABC} + \angle \mathrm{ADC} = 180^\circ = \angle \mathrm{ADC} + \angle \mathrm{CDE}. \] Subtracting the common \(\displaystyle \angle \mathrm{ADC}\) from both sides, \[\angle \mathrm{CDE} = \angle \mathrm{ABC}. \]Why the other choice of \(\displaystyle E\) gives the same thing. The two exterior angles at \(\displaystyle D\) — the one you get by producing \(\displaystyle AD\), and the one you get by producing \(\displaystyle CD\) — are vertically opposite angles, so they are equal. Each of them equals \(\displaystyle 180^\circ - \angle \mathrm{ADC}\), which is \(\displaystyle \angle \mathrm{ABC}\).The same argument works at every vertex, because at each vertex the interior angle has a linear pair with the exterior angle, and the interior angle is supplementary to the interior opposite angle.The exterior angle at any vertex and the interior angle at that same vertex add to \(\displaystyle 180^\circ\) (linear pair); the interior angle at that vertex and the interior opposite angle also add to \(\displaystyle 180^\circ\) (cyclic quadrilateral). Being supplements of the same angle, the exterior angle equals the interior opposite angle: \(\displaystyle \angle \mathrm{CDE} = \angle \mathrm{ABC}\).

Compare the chord with two radii — the triangle inequality.Let the circle have centre \(\displaystyle O\) and radius \(\displaystyle r\), and let \(\displaystyle PQ\) be any chord. Join \(\displaystyle OP\) and \(\displaystyle OQ\); both are radii, so \[OP = OQ = r. \] There are exactly two possibilities.Case $\displaystyle 1$: the chord passes through the centre. Then \(\displaystyle P, O, Q\) lie on one straight line and \[PQ = PO + OQ = r + r = 2r. \] This chord is a diameter, and its length is \(\displaystyle 2r\).Case $\displaystyle 2$: the chord does not pass through the centre. Then \(\displaystyle O, P, Q\) are three points not on one line, so \(\displaystyle OPQ\) is a genuine triangle. In any triangle, one side is shorter than the sum of the other two (the triangle inequality — you cannot get from \(\displaystyle P\) to \(\displaystyle Q\) more cheaply by going the straight way than by detouring through \(\displaystyle O\)). Hence \[PQ < OP + OQ = 2r. \]So every chord has length at most \(\displaystyle 2r\), and the length \(\displaystyle 2r\) is reached only in Case $\displaystyle 1$, when the chord is a diameter.Another way to see it. If a chord is at distance \(\displaystyle d\) from the centre, then the perpendicular from \(\displaystyle O\) bisects it and Pythagoras gives \[\text{length} = 2\sqrt{r^2 - d^2}. \] Since \(\displaystyle d \ge 0\), this is largest exactly when \(\displaystyle d = 0\), that is, when the chord passes through the centre, and then the length is \(\displaystyle 2r\).Every chord is at most \(\displaystyle 2r\) long, and a chord of length exactly \(\displaystyle 2r\) has to pass through the centre — so no chord is longer than a diameter, and the diameter is the longest chord.

The length of a chord is decided entirely by its distance from the centre.Let the circle have centre \(\displaystyle O\) and radius \(\displaystyle r\), and let \(\displaystyle A\) be a point inside it with \(\displaystyle OA = a\), where \(\displaystyle 0 < a < r\). (If \(\displaystyle A\) happens to be the centre, every chord through \(\displaystyle A\) is a diameter and they all have the same length \(\displaystyle 2r\); so take \(\displaystyle A \ne O\).)Step $\displaystyle 1$: a chord at distance \(\displaystyle d\) from the centre has length \(\displaystyle 2\sqrt{r^2 - d^2}\). Let \(\displaystyle PQ\) be a chord and let \(\displaystyle M\) be the foot of the perpendicular from \(\displaystyle O\) to \(\displaystyle PQ\); so \(\displaystyle OM = d\). The perpendicular from the centre bisects the chord, so \(\displaystyle PM = MQ\). In the right triangle \(\displaystyle OMP\), \[PM^2 = OP^2 - OM^2 = r^2 - d^2, \qquad PQ = 2PM = 2\sqrt{r^2 - d^2}. \] Read that formula carefully: the bigger \(\displaystyle d\) is, the shorter the chord. So finding the shortest chord through \(\displaystyle A\) is the same job as finding the chord through \(\displaystyle A\) that is farthest from the centre.Step $\displaystyle 2$: for a chord through \(\displaystyle A\), the distance \(\displaystyle d\) can never exceed \(\displaystyle OA\). Take any chord \(\displaystyle PQ\) passing through \(\displaystyle A\), and again let \(\displaystyle M\) be the foot of the perpendicular from \(\displaystyle O\) to \(\displaystyle PQ\). Both \(\displaystyle A\) and \(\displaystyle M\) lie on the line \(\displaystyle PQ\). So for every chord through \(\displaystyle A\), \(\displaystyle d \le a\), and \(\displaystyle d = a\) only for the chord perpendicular to \(\displaystyle OA\).Step $\displaystyle 3$: put the two steps together. Length \(\displaystyle =2\sqrt{r^2-d^2}\) gets smaller as \(\displaystyle d\) gets bigger, and \(\displaystyle d\) is biggest (equal to \(\displaystyle a\)) exactly for the chord perpendicular to \(\displaystyle OA\) at \(\displaystyle A\). Therefore that chord is the shortest, and its length is \[2\sqrt{r^2 - a^2}. \]Numerical check. With \(\displaystyle r = 5\) and \(\displaystyle OA = 3\), chords through \(\displaystyle A\) were measured at every direction from \(\displaystyle 0^\circ\) to \(\displaystyle 179^\circ\); the shortest came out at the direction perpendicular to \(\displaystyle OA\), with length \(\displaystyle 8 = 2\sqrt{25-9}\), while the longest was the diameter, \(\displaystyle 10\).The shortest chord through \(\displaystyle A\) is the one perpendicular to \(\displaystyle OA\) at \(\displaystyle A\); its length is \(\displaystyle 2\sqrt{r^2 - OA^2}\). (At the other extreme, the longest chord through \(\displaystyle A\) is the diameter through \(\displaystyle A\), of length \(\displaystyle 2r\).)

Join the centre to the third vertex — it splits the triangle into two isosceles triangles.(The figure that goes with this question shows a circle with a diameter \(\displaystyle AB\) through the centre \(\displaystyle O\), a third point \(\displaystyle C\) on the circle, and the triangle \(\displaystyle ACB\). The one extra line you need to draw on it is the radius \(\displaystyle OC\).)Let \(\displaystyle AB\) be a diameter of a circle with centre \(\displaystyle O\) and radius \(\displaystyle r\), and let \(\displaystyle C\) be any other point on the circle. Join \(\displaystyle OC\). Because \(\displaystyle A\), \(\displaystyle B\) and \(\displaystyle C\) are all on the circle, \[OA = OB = OC = r. \] So the radius \(\displaystyle OC\) cuts \(\displaystyle \triangle ACB\) into two isosceles triangles, \(\displaystyle \triangle OAC\) and \(\displaystyle \triangle OBC\).Step 1. In \(\displaystyle \triangle OAC\), \(\displaystyle OA = OC\), so the base angles are equal: \[\angle \mathrm{OAC} = \angle \mathrm{OCA} = x. \]Step 2. In \(\displaystyle \triangle OBC\), \(\displaystyle OB = OC\), so likewise \[\angle \mathrm{OBC} = \angle \mathrm{OCB} = y. \]Step 3. Now look at the whole triangle \(\displaystyle ACB\). Its angle at \(\displaystyle A\) is \(\displaystyle x\), its angle at \(\displaystyle B\) is \(\displaystyle y\), and its angle at \(\displaystyle C\) is made of the two pieces, \(\displaystyle \angle \mathrm{ACB} = x + y\). The three angles of a triangle add to \(\displaystyle 180^\circ\): \[x + y + (x+y) = 180^\circ \;\Longrightarrow\; 2(x+y) = 180^\circ \;\Longrightarrow\; x + y = 90^\circ. \]Hence \[\angle \mathrm{ACB} = x + y = 90^\circ. \]The same figure, seen a second way. \(\displaystyle AB\) is a diameter, so \(\displaystyle A\), \(\displaystyle O\), \(\displaystyle B\) are on a straight line and the angle at the centre standing on the arc \(\displaystyle AB\) is \(\displaystyle \angle \mathrm{AOB} = 180^\circ\). The angle at the centre is twice the angle at the circumference standing on the same arc, so \[\angle \mathrm{ACB} = \tfrac{1}{2}\angle \mathrm{AOB} = \tfrac{1}{2}\times 180^\circ = 90^\circ. \] The straight diameter is simply a "flat" central angle, and half of a straight angle is a right angle.Notice that nothing in the argument depended on where \(\displaystyle C\) was chosen — only on \(\displaystyle OC\) being a radius. So it works for every point of the semicircle.\(\displaystyle \angle \mathrm{ACB} = 90^\circ\): the angle in a semicircle is a right angle, because the radius \(\displaystyle OC\) splits the triangle into two isosceles triangles whose base angles are \(\displaystyle x\) and \(\displaystyle y\) with \(\displaystyle 2(x+y) = 180^\circ\).

Reflect the whole picture in the diameter \(\displaystyle AB\).Set-up: a circle with centre \(\displaystyle O\); \(\displaystyle AB\) a diameter; \(\displaystyle CC'\) and \(\displaystyle DD'\) two chords, each perpendicular to \(\displaystyle AB\), with \(\displaystyle C\) and \(\displaystyle D\) on one side of \(\displaystyle AB\) and \(\displaystyle C'\) and \(\displaystyle D'\) on the other. \(\displaystyle M\) is the midpoint of \(\displaystyle CD\) and \(\displaystyle M'\) is the midpoint of \(\displaystyle C'D'\).Step $\displaystyle 1$: \(\displaystyle AB\) is the perpendicular bisector of each of the two chords. The chord \(\displaystyle CC'\) is perpendicular to \(\displaystyle AB\), and \(\displaystyle AB\) passes through the centre \(\displaystyle O\). So the perpendicular dropped from \(\displaystyle O\) onto \(\displaystyle CC'\) runs along \(\displaystyle AB\) — and the perpendicular from the centre to a chord bisects that chord. Hence \(\displaystyle AB\) cuts \(\displaystyle CC'\) at right angles and into two equal halves; that is, \(\displaystyle AB\) is the perpendicular bisector of \(\displaystyle CC'\). In exactly the same way, \(\displaystyle AB\) is the perpendicular bisector of \(\displaystyle DD'\).Step $\displaystyle 2$: therefore \(\displaystyle C'\) is the mirror image of \(\displaystyle C\) in the line \(\displaystyle AB\), and \(\displaystyle D'\) of \(\displaystyle D\). Being the mirror image of a point in a line means precisely that the line is the perpendicular bisector of the segment joining the two points — which is what Step $\displaystyle 1$ established.Step $\displaystyle 3$: a mirror sends midpoints to midpoints. Reflection in \(\displaystyle AB\) keeps all lengths unchanged, so it carries the segment \(\displaystyle CD\) exactly onto the segment \(\displaystyle C'D'\). The point that sits halfway along \(\displaystyle CD\) must go to the point that sits halfway along \(\displaystyle C'D'\). So the mirror image of \(\displaystyle M\) is \(\displaystyle M'\).Step $\displaystyle 4$: finish. Since \(\displaystyle M\) and \(\displaystyle M'\) are mirror images of each other in the line \(\displaystyle AB\), the line \(\displaystyle AB\) is the perpendicular bisector of the segment \(\displaystyle MM'\). In particular \[MM' \perp AB, \] and, as a bonus, \(\displaystyle AB\) also bisects \(\displaystyle MM'\).A check with coordinates. Put \(\displaystyle O\) at the origin with \(\displaystyle AB\) along the \(\displaystyle x\)-axis, and let the radius be \(\displaystyle r\). A chord perpendicular to \(\displaystyle AB\) is a vertical segment. Say \(\displaystyle CC'\) sits at \(\displaystyle x = a\) and \(\displaystyle DD'\) at \(\displaystyle x = b\); then, using \(\displaystyle x^2+y^2=r^2\), \[C = \left(a,\; \sqrt{r^2-a^2}\right),\quad C' = \left(a,\; -\sqrt{r^2-a^2}\right),\quad D = \left(b,\; \sqrt{r^2-b^2}\right),\quad D' = \left(b,\; -\sqrt{r^2-b^2}\right). \] Writing \(\displaystyle h=\sqrt{r^2-a^2}\) and \(\displaystyle k=\sqrt{r^2-b^2}\), the midpoints are \[M = \left(\frac{a+b}{2},\; \frac{h+k}{2}\right), \qquad M' = \left(\frac{a+b}{2},\; -\frac{h+k}{2}\right). \] The two points have the same \(\displaystyle x\)-coordinate, so \(\displaystyle MM'\) is a vertical segment, and vertical means perpendicular to the \(\displaystyle x\)-axis, i.e. to \(\displaystyle AB\). (Worked out for \(\displaystyle r=6\) with several pairs \(\displaystyle (a,b)\), the two midpoints always shared their \(\displaystyle x\)-coordinate exactly.)A third route, if you prefer quadrilaterals. \(\displaystyle CC'\) and \(\displaystyle DD'\) are both perpendicular to \(\displaystyle AB\), so \(\displaystyle CC' \parallel DD'\). Then \(\displaystyle CDD'C'\) is a trapezium with \(\displaystyle CC'\) and \(\displaystyle DD'\) as its parallel sides, and \(\displaystyle CD\), \(\displaystyle C'D'\) as the other two sides. The segment joining the midpoints of those two sides is the mid-segment of the trapezium, which is parallel to the parallel sides. So \(\displaystyle MM' \parallel CC' \perp AB\).One degenerate case worth naming: if the points were labelled so that \(\displaystyle M\) and \(\displaystyle M'\) came out at the very same point on \(\displaystyle AB\), there would be no segment \(\displaystyle MM'\) to speak of. With the natural labelling above (\(\displaystyle C, D\) on one side of \(\displaystyle AB\) and \(\displaystyle C', D'\) on the other) this cannot happen.\(\displaystyle MM' \perp AB\) — indeed \(\displaystyle AB\) is the perpendicular bisector of \(\displaystyle MM'\), because reflection in the diameter \(\displaystyle AB\) interchanges \(\displaystyle C \leftrightarrow C'\), \(\displaystyle D \leftrightarrow D'\) and hence \(\displaystyle M \leftrightarrow M'\).

Use "the angle at the centre is twice the angle at the circumference" — twice, on both of the two arcs.(The figure that goes with this question shows a cyclic quadrilateral \(\displaystyle ABCD\) with the centre \(\displaystyle O\) marked, and the two radii \(\displaystyle OB\) and \(\displaystyle OD\) drawn — those two radii are the key to the argument.)Let \(\displaystyle ABCD\) be a quadrilateral with all four vertices on a circle of centre \(\displaystyle O\). Join \(\displaystyle OB\) and \(\displaystyle OD\).Step $\displaystyle 1$: the two radii create two angles at the centre that fill a complete turn. The chord \(\displaystyle BD\) divides the circle into two arcs: the arc \(\displaystyle BCD\) (the one containing \(\displaystyle C\)) and the arc \(\displaystyle BAD\) (the one containing \(\displaystyle A\)). Correspondingly the two radii \(\displaystyle OB\) and \(\displaystyle OD\) create two angles at \(\displaystyle O\): Together these two angles go all the way round the point \(\displaystyle O\), so \[p + q = 360^\circ. \] (One of the two will be a reflex angle — bigger than \(\displaystyle 180^\circ\) — and that is perfectly fine; the doubling rule below still holds.)Step $\displaystyle 2$: halve each one. \(\displaystyle \angle \mathrm{BAD}\) is an angle at the circumference standing on the arc \(\displaystyle BCD\) — the same arc as \(\displaystyle p\). So \[p = 2\,\angle \mathrm{BAD}. \] \(\displaystyle \angle \mathrm{BCD}\) is an angle at the circumference standing on the arc \(\displaystyle BAD\) — the same arc as \(\displaystyle q\). So \[q = 2\,\angle \mathrm{BCD}. \]Step $\displaystyle 3$: add. \[2\,\angle \mathrm{BAD} + 2\,\angle \mathrm{BCD} = p + q = 360^\circ, \] \[\angle \mathrm{BAD} + \angle \mathrm{BCD} = 180^\circ. \] That is the first pair of opposite angles.Step $\displaystyle 4$: the other pair comes free. The four angles of any quadrilateral add to \(\displaystyle 360^\circ\), so \[\angle \mathrm{ABC} + \angle \mathrm{ADC} = 360^\circ - \left(\angle \mathrm{BAD} + \angle \mathrm{BCD}\right) = 360^\circ - 180^\circ = 180^\circ. \] (You could also just repeat Steps $\displaystyle 1$–$\displaystyle 3$ with the radii \(\displaystyle OA\) and \(\displaystyle OC\) instead.)A quick sanity check. Take a rectangle, which is cyclic: each angle is \(\displaystyle 90^\circ\), and \(\displaystyle 90^\circ+90^\circ=180^\circ\). Take a square inscribed in a circle, same thing. And notice the statement fails for a non-cyclic quadrilateral — for instance a parallelogram that is not a rectangle has opposite angles equal but not \(\displaystyle 180^\circ\), which is exactly why a non-rectangular parallelogram can never be inscribed in a circle.Each pair of opposite angles adds to \(\displaystyle 180^\circ\): the two angles at the centre on the two arcs cut off by a diagonal add to \(\displaystyle 360^\circ\), and each opposite angle of the quadrilateral is half of one of them, so the pair adds to \(\displaystyle \tfrac{1}{2}\times 360^\circ = 180^\circ\).