Question

if A circle touches all the 4 sides of a quadrilateral ABCD at points P,Q,R,S then prove that AB+CD= BC +DA​

Answer 1

The figure shows that the tangents drawn from the exterior point to a circle are equal in length.

As DR and DS are tangents from exterior point D, DR = DS ...(1)

As AP and AS are tangents from exterior point A, AP = AS ...(2)

As BP and BQ are tangents from exterior point B, BP = BQ ...(3)

As CR and CQ are tangents from exterior point C, CR = CQ ...(4)

Adding (1), (2), (3) & (4), we get

DR + AP + BP + CR = DS + AS + BQ + CQ

⇒(DR+CR) + (AP+BP) = (DS+AS) + (BQ+CQ)

⇒CD + AB = DA + BC

⇒AB + CD = BC + DA

Hence, proved.

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