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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