# Given a square inscribed in a right triangle find the length of the hypotenuse

If $ABCD$ is a square, $OEF$ is a right triangle, $OA = 48$, and $OB = 36$, what is the length of the segment $EF$?

Well first I found that the length of a side of a square is $60$ by the Pythagorean theorem. Then I also get a lot of similar triangles such as $\triangle FOE \sim \triangle FDA$, $\triangle FOE \sim \triangle BC$E, $\triangle FDA \sim \triangle BCE$

I know that $FE= 60+FD+CE$ so I just need to find a way to relate what I know to $FD$ and $CE$. It seems straight forward but I just keep going in circles any help would be appreciated or if you know of a simpler approach.

Use $$\triangle FDA \sim \triangle AOB$$ to find $\frac{FD}{DA}$ and also use $$\triangle ECB \sim \triangle BOA$$
to find $\frac{EC}{CB}$.