4. On the side $AB$ of triangle $ABC$, a point $K$ is marked, and on the sides $AC$ and $BC$, points $L$ and $M$ are chosen respectively such that $AK = AL$ and $BK = BM$. It turns out that the lines $LM$ and $AB$ are parallel. The tangent at point $L$ to the circumcircle of triangle $KLM$ intersects the segment $CK$ at point $D$, and a line passing through $D$ and parallel to side $AB$ intersects side $BC$ at point $E$. Find the angle $\angle DEO$, where $O$ is the center of the circumcircle of triangle $KLM$.
Answer: $\angle D E O=90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-12.jpg?height=662&width=920&top_left_y=517&top_left_x=588)

First solution. By the problem's condition, $O K=O L=O M$ and triangles $A K O$ and $A L O$ are equal by three sides, so $\angle A O K=\angle A O L=\angle K M L$, the latter because $\angle K M L$ is an inscribed angle subtending the same arc as the central angle $\angle K O L$. From the parallelism of lines $A B$ and $L M$ and the isosceles nature of triangle $K B M$, it follows that $\angle K M L=\angle M K B=\angle K M B$. Therefore,

$\angle K O L=2 \angle A O K=2 \angle K M L=\angle K M L+\angle K M B=\angle B M L=180^{\circ}-\angle C M L=180^{\circ}-\angle B$.

Similarly, $\angle K O M=180^{\circ}-\angle A$. Therefore,

$$
\angle L O M=360^{\circ}-\angle K O L-\angle K O M=\angle A+\angle B=180^{\circ}-\angle C
$$

Thus, quadrilateral $C L O M$ is cyclic and $\angle C O L=\angle C M L$. Therefore,

$$
\angle C O K=\angle C O L+\angle K O L=\angle C M L+\left(180^{\circ}-\angle C M L\right)=180^{\circ}
$$

and, therefore, point $O$ lies on line $C K$.

Let $E^{\prime}$ be the intersection of the altitude of isosceles triangle $L O M$ with line $B C$. Then it lies on the perpendicular bisector of segment $L M$ and, in particular, $\angle E^{\prime} L M=\angle E^{\prime} M L=\angle B$. Therefore, $\angle L E^{\prime} O=90^{\circ}-\angle B$. On the other hand, $\angle L D O=90^{\circ}-\angle L O D=90^{\circ}-\angle B$. Thus, quadrilateral $O L D E^{\prime}$ is cyclic, so $\angle D E^{\prime} O=180^{\circ}-\angle D L O=90^{\circ}$. Therefore, $D E^{\prime}$ is parallel to $L M$ and, therefore, points $E$ and $E^{\prime}$ coincide.

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-13.jpg?height=711&width=916&top_left_y=158&top_left_x=587)

Second solution. By the problem's condition, triangles $K A L$ and $K B M$ are isosceles, so the perpendicular bisectors of segments $K L$ and $K M$, which intersect at point $O$, are the angle bisectors of angles $\angle A$ and $\angle B$. Therefore, point $O$ is the incenter of triangle $A B C$ and, therefore, $C O$ is the angle bisector of angle $\angle C$. From the parallelism of lines $A B$ and $L M$ and the isosceles nature of triangle $K B M$, it follows that $\angle K M L=\angle M K B=\angle K M B$. Therefore, $M K$ is the bisector of the exterior angle of triangle $C L M$. Similarly, $L K$ is also the bisector of the exterior angle of triangle $C L M$. Therefore, $K$ is the excenter of triangle $C L M$. Therefore, point $K$ lies on the angle bisector of angle $\angle L C M$, i.e., on line $C O$.

Since $\angle K O L$ is a central angle subtending the same arc as the inscribed angle $\angle K M L$, $\angle K O L=2 \angle K M L$. Therefore,

$$
\angle C O L=180^{\circ}-\angle K O L=180^{\circ}-2 \angle K M L=180^{\circ}-\angle B M L=\angle C M L
$$

Thus, triangles $C L O$ and $C D E$ are similar by two angles. Therefore, $\frac{C O}{C E}=$ $\frac{C L}{C D}$, and then triangles $C L D$ and $C O E$ are similar by angles and the ratios of the sides adjacent to them. Perform a rotational homothety by angle $\angle A C O$ with center at point $C$, under which point $O$ maps to point $L$. Then triangle $C O E$ maps to triangle $C L D$ and, in particular, line $O E$ maps to line $L D$. Therefore, the angle between these lines is $\angle A C O$. Denote their intersection point by $F$, then $\angle L F O=\angle L C O$ and, therefore, quadrilateral $C L O F$ is cyclic. Therefore, $\angle F C O=\angle F L O=90^{\circ}$. Similarly, $\angle D C E=\angle D F E$ and, therefore, quadrilateral $C D E F$ is cyclic. Therefore, $\angle D E F=\angle D C F=90^{\circ}$ and, therefore, $\angle D E F=90^{\circ}$.