In the following figure, OABC is a square. A circle is drawn with O as center which meets OC at P and OA at Q. Prove that :

(i) △ OPA ≅ △ OQC

(ii) △ BPC ≅ △ BQA

(i) In △ OPA and △ OQC,

⇒ OP = OQ (Radius of same circle)

⇒ ∠AOP = ∠COQ (Both equal to 90°)

⇒ OA = OC (Sides of square)

∴ △ OPA ≅ △ OQC (By S.A.S. axiom)

Hence, proved that △ OPA ≅ △ OQC.

(ii) Proved above,

⇒ OC = OA ………..(1)

⇒ OP = OQ ………..(2)

Subtracting equation (2) from (1),

⇒ OC - OP = OA - OQ

⇒ CP = QA

In △ BPC and △ BQA,

⇒ BC = BA (Sides of square)

⇒ ∠PCB = ∠QAB (Both equal to 90°)

⇒ CP = QA (Proved above)

∴ △ BPC ≅ △ BQA (By S.A.S. axiom)

Hence, proved that △ BPC ≅ △ BQA.