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Question
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
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Solution
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.
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