Consider Fig. 5.15. If is perpendicular to , is perpendicular to , and , show that .

Question

Homi · Accepted Answer

Understand the Question

The perpendicular drawn from the center of a circle to any chord bisects that chord.
Using radii drawn to the endpoints of the chords, we form two right-angled triangles ( $\triangle CEA$ and $\triangle CHG$ ).
By proving these two right-angled triangles congruent using the RHS congruence criterion, we establish that the half-chords are equal ( $AE = GH$ ), which proves that the entire chords are equal ( $AB = GF$ ).

Step 1 · Perpendicular from Center Bisects Chord

The perpendicular from the center of a circle to a chord bisects the chord. Diagram 1

Since $CE \perp AB$ , point $E$ is the midpoint of $AB$ : $AB = 2AE$

Since $CH \perp GF$ , point $H$ is the midpoint of $GF$ : $GF = 2GH$

Step 2 · Prove Congruence of $\triangle CEA$ and $\triangle CHG$

In right-angled triangles $\triangle CEA$ and $\triangle CHG$ :

By RHS congruence criterion: $\triangle CEA \cong \triangle CHG$

By corresponding parts of congruent triangles (CPCTC): $AE = GH$

Step 3 · Show $AB = GF$

From Step 1:

\begin{aligned} AB &= 2AE \\ GF &= 2GH \end{aligned}

Substituting $AE = GH$ into the expression for $AB$ :

\begin{aligned} AB &= 2AE \\ &= 2GH \\ &= GF \end{aligned}

$\therefore AB = GF$

Answer

Hence proved, $AB = GF$ .

Common Mistakes

Missing Chord Bisector Property: Forgetting that $CE \perp AB$ implies $AB = 2AE$ and $GF = 2GH$ .
Incorrect Congruence Criterion: Stating SAS or SSS instead of the RHS criterion, since the equality relies on the hypotenuse (radius) and one leg (given distance from center).

Question 2