Question

If a quadrilateral $ABCD$ is circumscribed about a circle with center $AB+CD=BC+DA$.

Answer 1

We know that the tangents drawn from the same external point are equal in length, Hence

$DR=DS$ (tangents drawn from the point $D$) $\to \left(1\right)$

$CR=CQ$ (tangents drawn from the point $C$) $\to \left(2\right)$

$BP=BQ$ (tangents drawn from the point $B$) $\to \left(3\right)$

$AP=AS$ (tangents drawn from the point $A$) $\to \left(4\right)$

Now, add the equations $\left(1\right)\left(2\right)\left(3\right)$ & $\left(4\right)$

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

$\Rightarrow CD+AB=AD+BC$