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Question
In the given figure P is a midpoint of chord AB of the circle O. prove that OP ^ AB.
Solution
Given:
In the figure, O is centre of the circle and ab is chord.
P is the mid-point of AB ⇒ AP = PB
To prove: OP ⊥ AB
Construction: Join OA and OB
Proof:
In ΔOAP and ΔOBP
OA = OB ....[radii of the same circle]
OP = OP ....[common]
AP = PB ....[given]
∴ By Side-Side-Side criterion of congruency,
ΔOAP ≅ ΔOBP
The corresponding parts of the congruent triangles are congruent.
∴ ∠OPA = ∠OPB
But ∠OPA + ∠OPB = 180° ....[linear pair]
∴ ∠OPA = ∠OPB = 90°
Hence OP ⊥ AB.
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