# A quadrilateral ABCD is drawn to circumscribe a circle (see fig.).

Question:

A quadrilateral ABCD is drawn to circumscribe a circle (see fig.).

Prove that AB + CD = AD + BC.

Solution:

In fig., we observe that

$\mathrm{AP}=\mathrm{AS}$   ...(1)

$\{\because \mathrm{AP}$ and $\mathrm{AS}$ are tangents to the circle drawn from the point $\mathrm{A}\}$

Similarly, $\quad B P=B Q \quad \ldots(2)$

$\mathrm{CR}=\mathrm{CQ} \quad \ldots(3)$

$\mathrm{DR}=\mathrm{DS} \quad \ldots(4)$

Adding (1), (2), (3), (4), we have

$(\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{DR})=(\mathrm{AS}+\mathrm{DS})+(\mathrm{BQ}+\mathrm{CQ})$

$\Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}$