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.

Draw your parallelogram with AB horizontal and labelled counterclockwise ABCD. Since CD = CB = r, the radius of the circle, your parallelogram is a rhombus of side r. Draw line CE and Notice that triangle CDE is isosceles with two equal sides r, angle CDE is $45^\circ$ and DCE is a right angle. So DE is $r\sqrt 2$ and BC = r. Areas are proportional to the square of the similar sides.

Originally Posted by Walagaster
Since CD = CB = r, the radius of the circle, your parallelogram is a rhombus of side r.
Why did I miss this...

Originally Posted by Walagaster
Areas are proportional to the square of the similar sides.
How do you get this?

Originally Posted by Monoxdifly
Why did I miss this...

How do you get this?
It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are $k$ times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?

Originally Posted by Walagaster
It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are $k$ times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?
Let's see...
$\displaystyle (\frac12×r×h) : (\frac12×r\sqrt2×kh)$
Wait, are you implying that $\displaystyle k=\sqrt2$?

Originally Posted by Walagaster
Areas are proportional to the square of the similar sides.
Originally Posted by Monoxdifly
How do you get this?
Originally Posted by Walagaster
It's true for general planar areas. It's easy enough to check for triangles. Suppose you have two similar triangles where the linear dimensions (including base and height) of one are k times the corresponding dimensions of the other. You can check for yourself how their areas compare. What is the ratio of the areas?
Originally Posted by Monoxdifly
Let's see...
$\displaystyle (\frac12×r×h) : (\frac12×r\sqrt2×kh)$
Wait, are you implying that $\displaystyle k=\sqrt2$?
Please don't mix up two posts. I said areas are proportional to the square of similar sides. You asked how I got that. I explained it in the last post above. Then I asked you to check it out. Please reply to the blue question so you understand it before trying to apply it to your problem.

From the ratio itself I got $\displaystyle k=\sqrt2$. Thus, the ratio itself is 1 : 2.

From the ratio itself I got $\displaystyle k=\sqrt2$. Thus, the ratio itself is 1 : 2.