#### 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.

#### 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 …………… (1)

In ΔBPC and ΔBQA,

BC = BA (Sides of the square)

∠PCB = ∠QAB (both 90°)

PC = QA (by (1))

By Side – Angle – Side criterion of congruence,

∴ ΔBPC ≅ ΔBQA (by SAS)

Is there an error in this question or solution?

Solution 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. Concept: Concept of Circles.