The two right triangles have equal radii as hypotenuses and a common perpendicular side. Therefore, they are congruent by the RHS Congruence Criterion. Hence, the chord is divided into two equal parts.
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 ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)
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