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

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.\(\displaystyle \mathrm{AB}=\mathrm{GF}\).