1989 AHSME Problems/Problem 7

Problem

In $\triangle ABC$ , $\angle A = 100^\circ$ , $\angle B = 50^\circ$ , $\angle C = 30^\circ$ , $\overline{AH}$ is an altitude, and $\overline{BM}$ is a median. Then $\angle MHC=$

$[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE);[/asy]$

$\textrm{(A)}\ 15^\circ\qquad\textrm{(B)}\ 22.5^\circ\qquad\textrm{(C)}\ 30^\circ\qquad\textrm{(D)}\ 40^\circ\qquad\textrm{(E)}\ 45^\circ$

Solution

We are told that $\overline{BM}$ is a median, so $\overline{AM}=\overline{MC}$ . Drop an altitude from $M$ to $\overline{HC}$ , adding point $N$ , and you can see that $\triangle ACH$ and $\triangle MCN$ are similar, implying $\overline{HN}=\overline{NC}$ , implying that $\triangle MNH$ and $\triangle MNC$ are similar, so $\angle MHC=\angle C=30^\circ$ .

$[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); draw((11,3)--(11,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); dot((11,0)); label("A", (6,6), N); label("B", (0,0), W); label("C", (16,0), E); label("H", (6,0), S); label("M", (11,3), NE); label("N", (11,0), S);[/asy]$ These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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1989 AHSME Problems/Problem 7

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