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Question
The given figure shows a circle with center O. P is mid-point of chord AB.
Show that OP is perpendicular to AB.
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Solution
Given: in the figure, O is center of the circle, and AB is chord. P is a point on AB such that AP = PB.
We need to prove that, 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 ...[By c.p.c.t]
But ∠OPA + ∠OPB = 180° ...[Linear pair]
∴ ∠OPA = ∠OPB = 90°
Hence, OP ⊥ AB.
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| 1 | PM = QM | 1 | ... |
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