Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 2.9 Junior Find the answer

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 $\angle MHC=$

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

Your answer

Multiple choice: answer with the letter of the option you want.

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 congruent, so $\angle MHC=\angle C=30^\circ$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.