In an acute triangle $K L M$, $V$ is the intersection of its altitudes and $X$ is the foot of the altitude on side $K L$. The bisector of angle $X V L$ is parallel to side $L M$ and angle $M K L$ has a size of $70^{\circ}$.

What are the sizes of angles $K L M$ and $K M L$?

(L. Hozová)

Hint. Visualize the situation described and look for congruent angles.
In a given triangle, we denote $Y, Z$ as the feet of the other altitudes and $N, O$ as the intersections of the angle bisector of $\angle XVL$ with the sides $KL, KM$, see the figure. In the figure, we also mark the congruent angles:

![](https://cdn.mathpix.com/cropped/2024_04_17_cc5184b1f392581ce2bcg-2.jpg?height=605&width=612&top_left_y=186&top_left_x=722)

In the right triangle $KZL$, we know the angles at vertices $K$ and $Z$, so the angle at vertex $L$ has a measure of $180^{\circ}-90^{\circ}-70^{\circ}=20^{\circ}$. Similarly, the angle at vertex $M$ in the right triangle $KXM$ is $20^{\circ}$. Therefore, angles $\angle K LZ$ and $\angle KMX$ are congruent.

Lines $NO$ and $LM$ are parallel, so angles $\angle MLZ, \angle LVN$, and $\angle OVZ$ are congruent (alternate and corresponding angles). For the same reason, angles $\angle LMX, \angle MVO$, and $\angle NVX$ are also congruent. Line $NO$ is the angle bisector of $\angle XVL$, so angles $\angle LVN$ and $\angle NVX$—and thus all the angles mentioned earlier—are congruent.

Therefore, angles $\angle KLM$ and $\angle KML$ are congruent. These are the interior angles of triangle $KLM$, whose third angle is known. The measures of angles $\angle KLM$ and $\angle KML$ are $\left(180^{\circ}-70^{\circ}\right): 2=55^{\circ}$.

Notes. The congruence of angles $\angle KLM$ and $\angle KML$ means that triangle $KLM$ is isosceles with base $LM$. This fact can be directly derived as follows:

Line $NO$ is parallel to side $LM$, so it is perpendicular to the altitude to this side. Since line $NO$ is the angle bisector of $\angle XVL$, angles $\angle NVX$ and $\angle OVZ$ are congruent, and since this line is perpendicular to the altitude $KY$, angles $\angle K VX$ and $\angle K VZ$ are also congruent. Triangles $K VX$ and $K VZ$ are right triangles and have congruent angles at vertex $V$, so they also have congruent angles at vertex $K$. Therefore, the altitude $KY$ is the angle bisector of $\angle LKM$, which means that triangle $KLM$ is isosceles with base $LM$.

The measures of specific angles can be determined in various ways, see, for example, the following reasoning:

Triangles $KZL$ and $VXL$ are right triangles and have a common angle at vertex $L$. The interior angles of these triangles at vertices $K$ and $V$ are therefore congruent and have a measure of $70^{\circ}$ (from the problem statement). Line $VN$ is the angle bisector of $\angle XVL$, so angle $\angle NVX$ has a measure of $35^{\circ} (70: 2=35)$. The measure of angle $\angle N$ in triangle $NVX$ is $55^{\circ} (90-35=55)$. Lines $VN$ and $LM$ are parallel, so angles $\angle XNV$ and $\angle KLM$ are congruent (corresponding angles). The measure of angle $\angle KLM$ is $55^{\circ}$.