Question 2

We will show that chords equidistant from the center are equal in length.

Step 1 — Bisected chords

The line from the center is perpendicular to the chord. It bisects the chord into two equal parts. CE is perpendicular to AB. So, E is the midpoint of AB. This means $AB = 2AE$.

CH is perpendicular to GF. So, H is the midpoint of GF. This means $GF = 2GH$.

Step 2 — Congruent triangles

Let's consider two triangles. These are $\triangle CEA$ and $\triangle CHG$. CA and CG are radii of the same circle. So, $CA = CG$. We are given that $CE = CH$. $\angle CEA$ is a right angle. $\angle CHG$ is a right angle. So, $\angle CEA = \angle CHG$. By RHS congruence rule, $\triangle CEA \cong \triangle CHG$. Corresponding parts of congruent triangles are equal. Therefore, $AE = GH$.

Step 3 — Equal chords

From Step 1, we know $AB = 2AE$. From Step 1, we know $GF = 2GH$. From Step 2, we found $AE = GH$. Let's substitute $GH$ for $AE$.

$AB = 2AE$

$AB = 2GH$

We also know $GF = 2GH$. Therefore,

$\boxed{AB = GF}$

Answer

(i) AB = GF is shown.