In fig 2 a quadrilateral ABCD is drawn to circumscribe a circle, with center O, in such a way that the sides AB, BC, CD, and DA touch the circle at the points P, Q, R, and S respectively. Prove that AB + CD = BC + DA
Given ABCD is quadrilateral P,Q,R,S are point of contact
To prove AB + CD = BC + DA
theorem -
tangents drawn from an external point are equal
Using the above theorem we can say
AP = AS --- (1)
BP = BQ --- (2)
CR = CQ --- (3)
OQ = OS --- (4)
Adding (1) , (2) , (3) , (4)