Solve the previous question using the Baudhāyana–Pythagoras theorem.

Question

Homi · Accepted Answer

Understand the Question

The perpendicular from the center of a circle to a chord bisects the chord, so $AE = \dfrac{AB}{2}$ and $GH = \dfrac{GF}{2}$ .
The distances of the chords from the center are given to be equal: $CE = CH$ .
Since $CA$ and $CG$ are radii of the same circle, $CA = CG$ .
We apply the Baudhāyana–Pythagoras theorem in the two right-angled triangles $\Delta CEA$ and $\Delta CHG$ to prove that chord $AB = GF$ .

Step 1 · Chord Bisection by Perpendicular from Center

Perpendicular from center $C$ to chord $AB$ bisects the chord: $AE = \dfrac{AB}{2}$

Similarly, perpendicular $CH$ to chord $GF$ bisects the chord: $GH = \dfrac{GF}{2}$ Diagram 1

Step 2 · Apply Baudhāyana–Pythagoras Theorem

In right-angled triangle $\text{CEA}$ (right-angled at $E$ ): $CA^2 = CE^2 + AE^2$

In right-angled triangle $\text{CHG}$ (right-angled at $H$ ): $CG^2 = CH^2 + GH^2$

Step 3 · Compare Triangles and Equate Segments

Since $CA$ and $CG$ are radii of the same circle: $CA = CG$

$CA^2 = CG^2$

Substituting from Step 2: $CE^2 + AE^2 = CH^2 + GH^2$

Given that $CE = CH$ , so $CE^2 = CH^2$ . Subtracting $CE^2$ from both sides: $AE^2 = GH^2$

Taking square root on both sides: $AE = GH$

Step 4 · Conclude Chord Equality

Substitute $AE = \dfrac{AB}{2}$ and $GH = \dfrac{GF}{2}$ into $AE = GH$ : $\dfrac{AB}{2} = \dfrac{GF}{2}$

Multiplying both sides by $2$ : $AB = GF$

Answer

Hence proved, $AB = GF$ .

Common Mistakes

Forgetting Chord Bisection: Overlooking the property that a perpendicular from the center to a chord bisects it ( $AE = \dfrac{AB}{2}$ and $GH = \dfrac{GF}{2}$ ).
Overlooking Radii Equality: Missing the fact that $CA = CG$ as both are radii of the same circle, which is crucial for equating the two Pythagoras expressions.

Question 3