Question 3

We need to prove that chords AB and GF are equal in length.

Step 1 — Chord bisection

Let's look at the given information. We have a circle with center C. Line segment CE is perpendicular to chord AB. This means E is the midpoint of AB. So, AE is half of AB.

$AE = \frac{AB}{2}$

Similarly, CH is perpendicular to chord GF. This means H is the midpoint of GF. So, GH is half of GF.

$GH = \frac{GF}{2}$

Step 2 — Apply Baudhāyana–Pythagoras theorem

Now, let's consider the right-angled triangles formed. Triangle CEA is a right-angled triangle at E. Using the Baudhāyana–Pythagoras theorem:

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

Triangle CHG is a right-angled triangle at H. Using the Baudhāyana–Pythagoras theorem:

$CG^2 = CH^2 + GH^2$

Step 3 — Compare the triangles

We know that CA and CG are radii of the same circle. Therefore, their lengths are equal.

$CA = CG$

Squaring both sides, we get:

$CA^2 = CG^2$

Now, let's substitute the expressions from Step 2:

$CE^2 + AE^2 = CH^2 + GH^2$

We are given that CE = CH. So, their squares are also equal:

$CE^2 = CH^2$

Subtracting CE² from both sides of the equation:

$AE^2 = GH^2$

Since AE and GH are lengths, they must be positive. Taking the square root of both sides:

$AE = GH$

Step 4 — Conclude the proof

From Step 1, we established that AE = AB/2 and GH = GF/2. Since AE = GH, we can substitute these values:

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

Multiplying both sides by 2:

$AB = GF$

Answer

We have proved that AB = GF.