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Question
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
Solution
Given: In the figure, O is the center of the circle, and AB is a chord. P is a point on AB such that AP=PB.
We need to prove that, AAP=BP
Construction: Join OA and OB
Proof:
In right triangles ΔOAP and ΔOBP
Hypotenuse OA=OB .....[ radii of the same circle ]
Side OP= OP ...[ common ]
∴ By Right Angle- Hypotenuse- Side criterion of congruency, ΔOAP ≅ ΔOBP
The corresponding parts of the congruent triangles are congruent.
∴ AP=BP .....[ by c.p.c.t ]
Hence proved.
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