Question -
Answer -
Given : The sides of a quadrilateral ABCD touch the circle at P, Q, R and S respectively
To prove : AB + CD = AP + BC
Proof : AP and AS are the tangents to the circle from A
AP = AS ….(i)
Similarly BP = BQ ……(ii)
CR = CQ ….(iii)
and DR = DS ….(iv)
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
=> (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
=> AB + CD = AD + BC
Hence proved.