# 1963 AHSME Problems/Problem 18

## Problem

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: $[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C); draw(B--C); draw(F--E--A); draw(unitcircle); label("B", B, SW); label("C", C, SE); label("A", A, W); label("E", E, S); label("U", U, NE); label("M", M, NE); label("F", F, N); //Credit to MSTang for the asymptote[/asy]$ $\textbf{(A)}\ \triangle EFA \qquad \textbf{(B)}\ \triangle EFC \qquad \textbf{(C)}\ \triangle ABM \qquad \textbf{(D)}\ \triangle ABU \qquad \textbf{(E)}\ \triangle FMC$

## Solution

Note that $\triangle EUM$ is a 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. $[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C); draw(B--C); draw(F--E--A); draw(unitcircle); label("B", B, SW); label("C", C, SE); label("A", A, W); label("E", E, S); label("U", U, NE); label("M", M, NE); label("F", F, N); draw("D",(0,0),NE); draw(B--(0,0)--C,dotted); draw(F--A); dot(A); dot(B); dot(C); dot((0,0)); dot(E); dot(F); dot(U); dot(M); //Credit to MSTang for the asymptote[/asy]$

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 $\boxed{\textbf{(A)}}$.

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