Question

The following figure shows a circle with center O.

If OP is perpendicular to AB, prove that AP = BP.

Answer 1

Given:

A circle with center O

AB is a chord of the circle

OP is perpendicular to AB

OP ⊥ AB

P is the foot of the perpendicular (the point where OP meets AB)

Join OA and OB.

In △OAP and △OBP,

⇒ OP = OP (Common side)

⇒ OA = OB (Radius of same circle)

⇒ ∠OPA = ∠OPB (Both equal to 90°)

∴ △ OAP ≅ △ OBP (By R.H.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AP = BP.