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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