smc

Note that $△EUM$ is a right triangle with one of the angles being $\angle FEA$. This leads to prediction that $△FEA$ is the similar triangle as it shares an angle, and to prove this, we need to show that $FE$ is a diameter. If we let $D$ be on $FE$ so that $FD=DB$, then by SAS Congruency, $△DMB\cong △DMC$, so $FD=DB=DC$. Since three points define a circle and point $D$ is equidistant from three points, $D$ is the center, so $FE$ is a diameter. Therefore, $\angle EAF$ is a right angle, and by AA Similarity, we can confirm that $△EFA\sim △EUM$. The answer is $\boxed{△EFA}$.