Question

See in the given dp
...and solve this problem !!​

Answer 1

kindly see in the attached pic.

glad help you

it helps you

thank you ☺️

Answer 2

Step-by-step explanation:

Given : OABC is a square.

A circle with centre O.

To Prove : (i) ∆OPA ≅ ∆OQC

(ii) ∆BPC ≅ ∆BQA

Proof : (i) In ∆OPA and ∆OQC,

OP = OQ (radii of same circle)

OA = OC (sides of square)

∠AOP = ∠COQ = 90°

∴ ∆OPA ≅ ∆OQC ( By SAS congruency )

(ii) Now, OP = OQ (radii of same circle)

...(A)

OC = OA (sides of square)

...(B)

Subtracting (A) from (B), we get

OC - OP = OA - OQ

CP = AQ ...(C)

In ∆BPC and ∆BQA,

BC = BA (sides of square)

∠PCB = ∠QAB = 90°

CP = AQ [ from (C) ]

∴ ∆BPC ≅ ∆BQA (By SAS congruency)