Evdokimov M.A.

In triangle $A B C A B=B C, \angle B=20^{\circ}$. Point $M$ on the base $A C$ is such that $A M: M C=1: 2$, point $H$ is the projection of $C$ on $B M$. Find the angle $A H B$.
Let's complete the triangle to a rhombus $ABCD$. Let $O$ be the center of the rhombus. Then the line $BM$ divides the median $AO$ of triangle $ABD$ in the ratio $2:1$. Therefore, this line is also a median, meaning it passes through the midpoint $K$ of segment $AD$.

Notice that points $O$ and $H$ lie on the circle with diameter $BC$. We can reason in different ways.

First method. $\angle KHO = \angle BCO = \angle KAO$. Therefore, quadrilateral $ANOK$ is cyclic, and $\angle AHK = \angle AOK = 80^\circ$, while $\angle AHB = 100^\circ$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-67.jpg?height=471&width=1221&top_left_y=622&top_left_x=430)

Second method. $MH \cdot MB = MO \cdot MC = MA^2$. Therefore, the specified circle is tangent to the line $AC$, from which $\angle AHB = 180^\circ - \angle BAC = 100^\circ$.

## Answer

$100^\circ$.

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