Question

If a quadrilateral ABCD circumscribes the circle. Prove that $AB+CD=\frac{1}{2}$ (Perimeter of ABCD)

Answer 1

Quadrilateral ABCD circumscribes the circle.
$\therefore$ AB, BC, CD and DA are tangents to the circle at the point of contact P,Q,R and S respectively.
As, we know that, the lengths of tangents drawn from an external point to a circle are equal.
$\Rightarrow$ AP=AS ....(1)
BP =BQ .....(2)
CR =CQ ....(3)
DR =DS ....(4)
Now, on adding (1), (2), (3) and (4), we get,
$\Rightarrow AP+BP+CR+DR=AS+BQ+CQ+DS\Rightarrow (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)\Rightarrow AB+CD=AD+BC$
Adding AB +CD on both sides
AB +CD +AB +CD =AD +BC +AB +CD
$\Rightarrow$ 2(AB+CD) =Perimeter of ABCD
$\Rightarrow AB+CD=\frac{1}{2}$ (Perimeter of ABCD)