
Problem 8.2. Given a $\triangle A B C$. Let $M$ be the midpoint of $A B$, $\angle C A B=15^{\circ}$ and $\angle A B C=30^{\circ}$.

a) Find $\angle A C M$.
b) Prove that $C M=\frac{A B \cdot B C}{2 A C}$.

Chavdar Lozanov


Solution: a) Let $A H \perp B C$. Then $\angle H A B=60^{\circ}$ and $A H=$ $\frac{A B}{2}=H M$. It follows from $\angle H A C=\angle H A B-\angle C A B=45^{\circ}$ that $A H=H C$. Thus $H M=H C$ and $\angle H C M=\frac{180^{\circ}-\angle M H C}{2}$. But $\angle M H C=\angle A H B-\angle A H M=90^{\circ}-60^{\circ}=30^{\circ}$. Therefore $\angle H C M=75^{\circ}$, so $\angle A C M=\angle H C M-\angle H C A=75^{\circ}-45^{\circ}=30^{\circ}$.

b) Let $S$ be the area of $\triangle A B C$. We know that $S=\frac{B C \cdot A H}{2}=$ $\frac{B C \cdot A B}{4}$. Since $C M$ is a median, we obtain $S_{A C M}=\frac{S}{2}$. If $M P \perp$ $A C$, then $\frac{S}{2}=\frac{A C \cdot P M}{2}$. It follows from $\triangle P M C$ that $P M=$ $\frac{1}{2} M C\left(\angle P C M=30^{\circ}\right)$. Therefore $\frac{S}{2}=\frac{A C \cdot M C}{4}$, so $\frac{A C \cdot M C}{2}=$ $\frac{B C \cdot A B}{4}$, which implies $M C=\frac{A B \cdot B C}{2 A C}$.
