CD is height of right-angled triangle ABC, M and N are midpoints of CD and BD: prove AM⊥CN

Question

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): https://i.stack.imgur.com/VWLZl.jpg

Sorry for my English. Any help/hints appreciated.

Answer 1

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 $AD/CD=CD/BD$.

Answer 2

One way would be analytic geometry (but I am not sure it is the simplest): endow $\mathbb{R^2}$ with a coordinate system where $C=(0,0)$, $A=(0,a)$, $B=(b,0)$. The line $L(A,B)$ ($L(P,Q)$ is the line through $P$ and $Q$) is $y=-\frac{a}{b}x+a$, the line $L(C,D)$ is $y=\frac{b}{a}x$, then solve the coordinates of $D,M,N$ write the equation of lines $L(C,N)$ and $L(A,M)$, and check orthogonality...