Quadrilateral ABCD is an isosceles trapezoid, with bases AB and CD. A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base AB is 2x and the length of base CD is 2y. Prove that the radius of the inscribed circle is √xy .
If x = y, then the quadrilateral is a square, and r = sqrt(xy) = x. So it's true!
Drop a vertical from A meeting DC at E.
DE = (2y - 2x)/2 = y-x.
AD = x + y.
AE = 2R.
Now Pythagoras, AD^2 = AE^2 + DE^2, etc..