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?                                                                                                           [Page No. 101]

Let C be the centre and AB be a chord.

Let CM ⊥ AB with M on AB.

We need to show that AM = BM.

In triangles CMA and CMB,

CA = CB [Radii of the circle]

CM = CM [Common side]

∠CMA = ∠CMB = 90° [Given]

By RHS congruence, ΔCMA ≅ ΔCMB.

Hence, AM = BM [CPCT]

So, M is the midpoint of AB.

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

Hence, proved.

Caution
Students should remember that the perpendicular from the centre of a circle to a chord bisects the chord; they should not assume that just any line drawn from the centre bisects a chord.