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

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 \(\displaystyle \tfrac{1}{2}\times 360^\circ = 180^\circ\).