A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D. AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ....

Here I got that triangle BCH and triangle EHD is similar with angle BCH = angle HDE = 45°, angle CHB = angle DHE = 112.5°, and angle CBH = angle DEH = 22.5°. The area of triangle BCH is ½ × CH × h, where h is the parallelogram's height. The area of triangle EHD is ½ × (r - CH) × r. I stuck at the ratio is (CH × h) : ((r - CH) × r). How do I simplify that? Problem is, I don't know CD's length, so I can't approximate the ratio between CH and DH.