Question

Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle

Answer 1

Let MN be the diameter and chord AB of circle C (O, r) , then according to the question AP = BP.
Then we have to prove that ∠AOM = ∠BOM.


Join OA and OB
In ΔAOP and ΔBOP
OA =OB (Radii of the same circle)
AP = BP (P is the mid point of chord AB)
OP = OP (Common)
Therefore, $△$ AOP is congruent to $△$ BOP
∠AOP = ∠BOP (CPCT)
∠AOM = ∠BOM (CPCT)
Hence proved.