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

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.