The area an isosceles trapezoid is equal to $S$, and the height is equal to the half of one of the non-parallel sides. If a circle can be inscribed in the trapezoid, find, with the proof, the radius of the inscribed circle. Express your answer in terms of $S$ only.
I labeled the trapezoid $ABCD$ starting lower left corner going clockwise. The area $S$ is equal to $h\times\left({a+b\over 2}\right)$. So $S=\left({AD+BC\over 2}\right)\times\left({AB\over 2}\right)=\left({AB\times(AD+BC)\over 4}\right)$. I know intuitively that because the circle is inscribed and the tangents are parallel, the two perpendicular radii form the diameter, but I don't know how to prove that (I need to). From there its the same as the height I would guess, unsure how to proceed.