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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Let CM be perpendicular to chord AB. The triangles CMA and CMB are right triangles. Their hypotenuses, CA and CB, are equal because they are radii of the same circle, and CM is common. Hence, the triangles are congruent by the RHS Congruence Criterion. Therefore, AM = BM, proving that the perpendicular from the centre bisects the chord.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/