Question

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 $AB+CD=BC+DA$.

Answer 1

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 $DR=DS$. [Tangents from same point outside the circle are equal.]

On adding we get

$AP+BP+CR+DR=AS+BQ+CQ+DS$

$AB+CD=AD+BC$