8.4. In triangle $A B C$, angle $A$ is three times smaller than angle $C$, and side $B C$ is half the length of side $A B$. Find the angles of triangle $A B C$.
Answer. Angles $A, B$ and $C$ are equal to $30^{\circ}, 60^{\circ}, 90^{\circ}$ respectively. Solution. Let $B C=a, \angle A=\alpha$, then $A B=2 a$ and $\angle C=3 \alpha$. Take a point $M$ on side $A B$ such that $\angle A C M=\alpha$, then $\angle B M C=2 \alpha$ by the property of the exterior angle. Thus, $\angle B C M=3 \alpha-\alpha=2 \alpha$ and therefore, triangle $B M C$ is isosceles. Then $B M=a$ and $A M=2 a-a=a$, and hence $C M=A M=a$, since triangle $A M C$ is isosceles. Therefore, triangle $B M C$ is equilateral and thus, all its angles are equal to $60^{\circ}$ ( $=2 \alpha)$. The other angles can now be easily determined.