8.2. In triangle $A B C$, angle $A B C$ is 120 degrees, and points $K$ and $M$ are marked on side $A C$ such that $A K=A B$ and $C M = C B$. A perpendicular $K H$ is drawn from point $K$ to line $B M$. Find the ratio $B K : K H$.
Answer: $2: 1$.

Comments. Only the correct answer - $\underline{0 \text { points. }}$.

Solution. Since $A K=A B$ and $C M=C B$, then angle $K B M=$ angle $K B A+$ angle $M B C-$ angle $A B C=\left(180^{\circ}-\right.$ angle $\left.C A B\right) / 2+$ $\left(180^{\circ}-\right.$ angle $\left.A C B\right) / 2-120^{\circ}=30^{\circ}$. Since in the right triangle ВНК the angle at vertex В is $30^{\circ}$, then $B K: K H=2: 1$.