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Question
In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q.
Prove that:
( i ) ΔOPA ≅ ΔOQC
( ii ) ΔBPC ≅ ΔBQA
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Solution
(i) In ΔOPA and ΔOQC,
OP = OQ ....[ radii of same circle ]
∠AOP = ∠COQ ... [ both 90° ]
OA = OC ... [ sides of the square ]
By Side- Angle - Side criterion of congruence.
∴ ΔOPA ≅ ΔOQC ...[ by SAS ]
(ii) Now, OP = OQ ...[ radii ]
and OC = OA ...[ sides of the square ]
∴ OC - OP = OA - OQ
⇒ CP = AQ ....(i)
In ΔBPC and ΔBQA,
BC = BA ...[ sides of the square ]
∠PCB = ∠QAB ...[ both 90° ]
PC = QA ...[ by ( i ) ]
By Side- Angle-Side criterion of congruence,
∴ ΔBPC ≅ ΔBQA ...[ by SAS ]
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