CD is height of right-angled triangle ABC, M and N are midpoints of CD and BD: prove AM⊥CN I was having some troubles proving this:
CD is the height that corresponds to the hypotenuse of right-angled triangle ABC. If M and N are midpoints of CD and BD, prove that AM is perpendicular to CN.
Here's the illustration (by me): http://i.imgur.com/JyR9Gz0.jpg
Sorry for my English. Any help/hints appreciated.

It suffices to show $\angle MAD=\angle NCD$, which means two right triangles 
$ADM$ and $CDN$ are similar. This can be proved from $AD/DM=CD/DN$, which follows from $AD/CD=CD/BD$.
