In the isosceles triangle $ABC$ with base $AC$, the orthocenter is $M$. We know that $AC = BM$. What are the angles of the triangle?

---

The isosceles triangle $ABC$ has base $AC$, and its orthocenter is $M$. Given that $AC = BM$, we need to determine the angles of the triangle.
Solution. First, let's calculate the angles in an acute triangle. In the isosceles triangle $ABC$ ($AB = AC$), let the angle at vertex $C$ be $2\beta$; the foot of the altitude from $B$ to $M$ is $B_1$, and the foot of the altitude from $C$ to $M$ is $C_1$.

![](https://cdn.mathpix.com/cropped/2024_05_02_12124dc2aac098e7759cg-1.jpg?height=445&width=371&top_left_y=271&top_left_x=866)

The triangle $B_1BA$ is similar to the triangle $C_1CA$. Both are right triangles and the angle at vertex $A$ is common to both. This implies that $\angle C_1CA = \beta$.

The triangles $C_1BM$ and $C_1CA$ are congruent: both are right triangles, $\angle C_1BM = \angle C_1CA = \beta$, and $BM = CA$ by the given condition. Due to the congruence, $BC_1 = C_1C$. Therefore, the triangle $BCC_1$ is isosceles, so $\angle C_1CB = 2\beta$.

The angles of the original triangle $ABC$ can be expressed in terms of $\beta$, thus

$$
2\beta + 2(2\beta + \beta) = 180^\circ, \quad \text{so} \quad 8\beta = 180^\circ, \quad \text{hence} \quad 2\beta = 45^\circ
$$

The other two angles of the triangle are:

$$
\frac{180^\circ - 45^\circ}{2} = \frac{135^\circ}{2} = 67.5^\circ
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_12124dc2aac098e7759cg-1.jpg?height=464&width=374&top_left_y=1275&top_left_x=865)

Now, let the triangle $ABC$ be obtuse. In this case, the orthocenter lies outside the triangle, and $B$ and $M$ switch roles. We previously saw that $\angle CMA = 45^\circ$ and $\angle MCA = 67.5^\circ$. Following the steps of the previous proof, we can also show that the triangle $CC_1M$ is isosceles. Since $\angle C_1CM = 45^\circ$, we get that $\angle BCA = 67.5^\circ - 45^\circ = 22.5^\circ$ and $\angle ABC = 135^\circ$.

Thus, the angles of the triangle $ABC$ are: $22.5^\circ, 135^\circ$, and $22.5^\circ$.