Circle In A Quadrilateral

By the property of tangent lines to circles, we have:

$\lvert\overline{AP}\rvert = \lvert\overline{AS}\rvert, \lvert\overline{BP}\rvert = \lvert\overline{BQ}\rvert, \lvert\overline{CQ}\rvert = \lvert\overline{CR}\rvert, \lvert\overline{DR}\rvert = \lvert\overline{DS}\rvert.$

Thus, we must have $AD + BC = AB + DC$ ; in other words, the opposite sides must sum to the same length. Thus, $AD + 6 = 8 + 5,$ so $AD=7.$

Let $x$ be the missing side.

By the pitot's theorem , we have

$6+x=5+8$

$6+x=13$

$x=13-6=$ $\boxed{7}$

Let X + Y = 5, and M + Y = 6. Then it becomes obvious that Y = 1. Then the rest follows