• # question_answer 7) In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre 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 $AB+CD=BC+DA$.

 We have, AB, BC, CD and DA are the tangents touching the circle at P, Q, R and S respectively Now, $AP=AS,BP=BQ,CR=CQ$ and $DR=DS$. On adding we get $AP+BP+CR+DR=AS+BQ+CQ+DS$ $AB+CD=AD+BC$                         Hence Proved.