Question

In the following figure,OABC is a square.A circle is drawn with O as centre which centre which meets OC at P and OA at Q.Prove that:
(i)∆OPA congruent to ∆OQC
(ii)∆BPC congruent to ∆BQA
pls give right answers no false answers
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Answer 1

║⊕ANSWER⊕║

i) In ΔOPA and ΔOQC,

S    OP = OQ (radii of same circle)

A    ∠AOP = ∠COQ (both 90°)

S    OA = OC (Sides of the square)

By Side – Angle – Side criterion of congruence  ΔOPA ≅ ΔOQC (by SAS congruency criterion)

(ii)

Now, OP = OQ (radii)

And OC = OA (sides of the square)

∴ OC – OP = OA – OQ ⇒ CP = AQ …………… (1)

In ΔBPC and ΔBQA,

S   BC = BA (Sides of the square)

A  ∠PCB = ∠QAB (both 90°)

S   PC = QA (by (1))

By Side – Angle – Side criterion of congruence,

∴ ΔBPC ≅ ΔBQA (by SAS congruency criterion)

HENCE PROVED !!!!!

Answer 2

Answer:

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Step-by-step explanation:

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