Q. 385.0( 5 Votes )

A qua

Answer :

We have

Given: ABCD is a quadrilateral and it circumscribe a circle that touches the quadrilateral at P, Q, R, S.

To Prove: AB + CD = AD + BC

Proof: Since tangents from an external point to the circle are equal, we can write

AP = AS …(i)

BP = BQ …(ii)

DR = DS …(iii)

CR = CQ …(iv)

Adding (i), (ii), (iii) and (iv), we get

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

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

AB + CD = AD + BC

Hence, proved.

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