8,9

Angle $A$ at the vertex of isosceles triangle $A B C$ is $100^{\circ}$. On ray $A B$, segment $A M$ is laid off, equal to the base $B C$. Find angle $B C M$.
On the ray $B M$, we lay off the segment $B N$ equal to $A M$. Let $A C = A B = a, B N = A M = B C = b$. The angles at the base $C N$ of the isosceles triangle $C B N$ are each $20^{\circ}$. Then

$$
B M = A M - A B = b - a, \quad M N = B N - B M = b - (b - a) = a, \quad C N = 2 B C \cos 20^{\circ} = 2 b \cos 20^{\circ},
$$

$$
\frac{B M}{M N} = \frac{b - a}{a} = \frac{b}{a} - 1 = \frac{\sin 100^{\circ}}{\sin 40^{\circ}} - 1, \quad \frac{C B}{C N} = \frac{b}{2 b \cos 20^{\circ}} = \frac{1}{2 \cos 20^{\circ}}.
$$

We will prove that $\frac{B M}{M N} = \frac{C B}{C N}$. From this it will follow that $C M$ is the angle bisector of triangle $B C N$, and then

$$
\angle B C M = \frac{1}{2} \angle B C N = \frac{1}{2} \cdot 20^{\circ} = 10^{\circ}
$$

Indeed,

$$
\begin{gathered}
\frac{B M}{M N} = \frac{C B}{C N} \Leftrightarrow \frac{\sin 100^{\circ}}{\sin 40^{\circ}} - 1 = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \\
\Leftrightarrow \frac{\sin 100^{\circ} - \sin 40^{\circ}}{\sin 40^{\circ}} = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \frac{2 \sin 30^{\circ} \cos 70^{\circ}}{2 \sin 20^{\circ} \cos 20^{\circ}} = \frac{1}{2 \cos 20^{\circ}} \Leftrightarrow \\
\Leftrightarrow \frac{\cos 70^{\circ}}{\sin 20^{\circ}} = 1 \Leftrightarrow \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1
\end{gathered}
$$

Which is what we needed to prove.

## Answer

$10^{\circ}$.

It is known that for some internal point $K$ of the median $B M$ of triangle $A B C$, the angles $B A K$ and $B C K$ are equal. Prove that triangle $A B C$ is isosceles.

## Solution

It is sufficient to prove that $B K \perp A C$. Suppose this is not the case. Let $\angle B M C < 90^{\circ}$, and let $T$ be the point symmetric to vertex $C$ with respect to the line $B M$. Then triangle $A M T$ is isosceles, and $M B$ is the bisector of its external angle at the vertex, so $A T \parallel B K$ and $\angle B T K = \angle B C K = \angle B A K$.

From points $A$ and $T$, lying on the same side of the line $B K$, the base $B K$ of the trapezoid $A T B K$ is seen at the same angle, meaning that this trapezoid is inscribed, and therefore, isosceles. The perpendicular bisector of its base $A T$ passes through the midpoint of the base $B K$, which is impossible, since the perpendicular bisector of the base $A T$ of the isosceles triangle $A M T$ passes through the point $M$, distinct from $K$.

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