The problem is
In △ABC, it is given that $\operatorname{m}(\measuredangle ACB) = 90^\circ$ and $AC = BC$. Let $M$ be the midpoint of $AB$. Prove that $CM < AM$.
I know that you can prove this by contradiction, and I've already proved that $CM$ can't equal $AM$. I just need help proving that $CM$ is not greater than $AM$. I've drawn a diagram as a reference. This is hyperbolic geometry, by the way.
