Solve the previous question using the Baudhāyana-Pythagoras theorem.                                                  [Page No. 104]

Given: E and H are midpoints of AB and GF, so

$$AE = \frac{AB}{2} \text{ and } FH = \frac{GF}{2}$$

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

$$CA^2 = CE^2 + AE^2$$

$$\Rightarrow AE^2 = CA^2 - CE^2 = r^2 - CE^2 \quad \dots(i)$$

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

$$CF^2 = CH^2 + FH^2$$

$$\Rightarrow FH^2 = r^2 - CH^2 \quad \dots(ii)$$

Since, CE = CH

From (i) and (ii),

$$AE^2 = FH^2$$

$$\Rightarrow AE = FH$$

$$\Rightarrow \frac{AB}{2} = \frac{GF}{2}$$

Therefore, AB = GF

Hence, proved.