Circles and Geometric Shapes | Exercise 5.3

Question 1

Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?

(Hint: Use Fig. 5.12. You are told that $\angle CMA = \angle CMB = 90^\circ$ . You need to show that $AM = BM$ .)

Solution

Understand the Question

We are given a circle with centre $C$ , a chord $AB$ , and a line segment $CM$ drawn from the centre perpendicular to the chord ( $CM \perp AB$ , so $\angle CMA = \angle CMB = 90^\circ$ ).
To prove that $CM$ bisects $AB$ , we need to show that $AM = BM$ .
We join radii $CA$ and $CB$ to form two right triangles, $\triangle CMA$ and $\triangle CMB$ , and prove they are congruent using the RHS (Right angle-Hypotenuse-Side) congruence criterion.

Step 1 · Identify Given Information and Setup

Consider a circle with centre $C$ and chord $AB$ . Diagram 1

Given: $CM \perp AB \implies \angle CMA = \angle CMB = 90^\circ$

To prove: $AM = BM$

Step 2 · Prove Congruence of Triangles

Join $CA$ and $CB$ .

In right triangles $\triangle CMA$ and $\triangle CMB$ :

$\angle CMA = \angle CMB = 90^\circ$ (Given)
$CA = CB$ (Radii of the same circle — Hypotenuse)
$CM = CM$ (Common side)

By the RHS congruence criterion: $\triangle CMA \cong \triangle CMB$

Therefore, by CPCT (Corresponding Parts of Congruent Triangles): $AM = BM$

Hence, the perpendicular from the centre to a chord bisects the chord.

Answer

Since $\triangle CMA \cong \triangle CMB$ by the RHS criterion, $AM = BM$ by CPCT. Therefore, the perpendicular from the centre bisects the chord.

Common Mistakes

Confusing the Theorem and its Converse: The original theorem uses SSS congruence (given midpoint, prove perpendicular), while the converse uses RHS congruence (given perpendicular, prove midpoint/bisects).
Using SAS Instead of RHS: Stating SAS congruence is incorrect because the angle ( $90^\circ$ ) is not the included angle between the radius and the common side.

Question 1

More questions in Exercise 5.3