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**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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