Problem 3. In a right triangle $ABC$ with a right angle at $C$, the bisector $BD$ and the altitude $CH$ are drawn. A perpendicular $CK$ is dropped from vertex $C$ to the bisector $BD$. Find the angle $HCK$, if $BK: KD=3: 1$.
Answer: $30^{\circ}$.

Solution. Let $M$ be the midpoint of $B D$. Then $C M$ is the median of the right triangle $C B D$ and $C M=M B=M D$. In addition, $C K$ is the height and median of triangle $M C D$, so $M C=C D$ and triangle $C M D$ is equilateral. Then $\angle C D M=60^{\circ}$ and $\angle C B M=$ $\angle C B D=90^{\circ}-60^{\circ}=30^{\circ}$, therefore, $\angle C B A=2 \angle C B D=60^{\circ}$. Hence $\angle B C H=30^{\circ}$, $\angle H C K=90^{\circ}-30^{\circ}-30^{\circ}=30^{\circ}$.

Answer to variant 2: $30^{\circ}$.