Use the Baudhāyana-Pythagoras theorem to show theorem: Chords of a circle having the same length are all at the same distance from the centre of the circle must be true.                                                                               [Page No. 104]

Theorem: Chords of a circle having the same length are all at the same distance from the centre of the circle.

Let AB and FG be two chords of a circle with centre C, such that AB = FG. Let E and H be the feet of perpendiculars from C to AB and FG respectively.

Since, the perpendicular from the centre of a circle to a chord of the circle bisects the chord.
E and H are midpoints of AB and FG.

$$\text{So, } AE = \frac{AB}{2} \text{ and } FH = \frac{FG}{2}$$

Since AB = FG, we get AE = FH.

Also, CA = CF = $r$ (radii of the circle).

In the right triangle CEA (right-angled at E), by the Baudhāyana–Pythagoras theorem:

$$CE^2 = CA^2 - AE^2 = r^2 - \left(\frac{AB}{2}\right)^2$$

Similarly, in right triangle CHF:

$$CH^2 = CF^2 - FH^2 = r^2 - \left(\frac{FG}{2}\right)^2$$

Since AB = FG, we have $CE^2 = CH^2$, and therefore CE = CH.

Hence, equal chords are equidistant from the centre.

Concept Applied
Equal chords of a circle are equidistant from the centre, and chords equidistant from the centre are equal (proved using the Baudhāyana–Pythagoras theorem).