
Problem 8'2. Let $M$ be the midpoint of the side $B C$ of $\triangle A B C$ and $\angle C A B=45^{\circ} ; \angle A B C=30^{\circ}$.

a) Find $\angle A M C$.

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

Chavdar Lozanov


Solution: a) Draw $C H \perp A B$. Now $\angle A C H=45^{\circ}=\angle C A H$ and $\angle H C B=60^{\circ}$. For $\triangle A C H$ it is true that $A H=H C$. Further it follows from $\triangle C H B$ that $C H=\frac{1}{2} C B=H M$. Therefore $A H=$ $H M$, so $\angle M A H=\angle A M H=\frac{180^{\circ}-\angle A H M}{2}$. Note that $\angle C H M=$ $60^{\circ}$ and $\angle A H M=90^{\circ}+60^{\circ}=150^{\circ}$. Therefore $\angle A M H=15^{\circ}$. We obtain that $\angle A M C=\angle H M C-\angle A M H=60^{\circ}-15^{\circ}=45^{\circ}$.

b) Let $S$ be the area of $\triangle A B C$. We know that $S=\frac{A B \cdot C H}{2}=$ $\frac{A B \cdot C B}{4}$. Since $A M$ is a median, it follows that $S_{A M C}=\frac{S}{2}$. If $C P \perp A M$, then $\frac{S}{2}=\frac{A M \cdot C P}{2}$. But $\angle C A M=\angle C A B-\angle M A B=$ $45^{\circ}-15^{\circ}=30^{\circ}$, and thus $C P=\frac{A C}{2}$. Therefore $\frac{S}{2}=\frac{A M \cdot A C}{4}$. Now $\frac{A B \cdot C B}{4}=\frac{A M \cdot A C}{2}$, and we obtain $A M=\frac{A B \cdot B C}{2 A C}$.
