Question

If the sides of a quadrilateral ABCD touch a circle, then prove that AB+CD=BC+AD

Answer 1

Given - A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q, R, and S respectively.

To Prove - AB+ CD = BC+AD
Proof -
By Theorem- Tangents drawn from an external point to a circle are equal in lengths.
$\because$ AP and AS are the tangents to the circle from external points A.
$\therefore$ AP=AS ....(i)
Similarly, we can prove that,
BP = BQ ....(ii)
CR=CQ ....(iii)
DR = DS ....(iv)

On adding (i), (ii), (iii) and (iv), we get:
AP+BP+CR+DR=AS+BQ+CQ+DS
$\Rightarrow$ [(AP+BP)+(CR+DR)] = [(AS+DS)+(BQ+CQ)]
$\Rightarrow$ AB+CD=AD+BC

Hence, AB+CD=BC+AD