Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. 
Between $B$ and $M$ point $U$ is taken, 
and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, $\triangle EUM$ is similar to:

$\textbf{(A)}\ \triangle EFA \qquad \textbf{(B)}\ \triangle EFC \qquad \textbf{(C)}\ \triangle ABM \qquad \textbf{(D)}\ \triangle ABU \qquad \textbf{(E)}\ \triangle FMC$
Note that $\triangle EUM$ is a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) with one of the angles being $\angle FEA$.  This leads to prediction that $\triangle FEA$ 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, $\triangle DMB \cong \triangle DMC$, 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 $\triangle EFA \sim \triangle EUM$.  The answer is $\boxed{\textbf{(A)}}$.