4. On the leg $AC$ of the right triangle $ABC$ with hypotenuse $AB$, a point $P$ is marked. Point $D$ is the foot of the perpendicular dropped from vertex $A$ to the line $BP$, and point $E$ is the foot of the perpendicular dropped from point $P$ to the side $AB$. On the plane, a point $T$ is chosen such that the lines $TA$ and $TP$ are tangents to the circumcircle of triangle $PAB$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the line $DE$ intersects the line $BC$ at point $Q$, and the line passing through point $C$ parallel to $OQ$ intersects the segment $BO$ at point $K$. Find the angle $\angle OKE$.
Answer: $90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-17.jpg?height=1031&width=1200&top_left_y=1266&top_left_x=428)

Solution. Let $\angle A B P=\varphi$ and $\angle C B P=\psi$. Triangles $A O T$ and $P O T$ are equal by three sides, and the inscribed angle $\angle A B P$ subtends the same arc as the central angle $\angle A O P$, so $\gamma=\angle A B P=\frac{1}{2} \angle A O P=\angle A O T$. Therefore, $\frac{O A}{O T}=\cos \varphi$. The central angle $\angle A O B$ subtends the smaller arc $A B$, while the inscribed angle $\angle A P B$ subtends the larger arc $A B$, so $\angle A P B+\frac{1}{2} \angle A O B=180^{\circ}$. From the isosceles triangle $A O B$, we have

$$
\angle A B O=90^{\circ}-\frac{1}{2} \angle A O B=90^{\circ}-\left(180^{\circ}-\angle A P B\right)=\angle A P B-90^{\circ}=\angle P B C=\psi
$$

Since $\angle A D P=90^{\circ}=\angle A E P$, quadrilateral $A D P E$ is cyclic, in particular, $\angle E A P=\angle E D P=\angle B D E$. Therefore,

$$
\angle B D E+\angle D B O=\angle B D E+\angle D B A+\angle A B O=\angle E A P+\angle A O T+\angle B A O=90^{\circ}
$$

Thus, lines $D E$ and $B O$ are perpendicular, and lines $B O$ and $Q T$ are parallel. Points $O$ and $T$ lie on the perpendicular bisector of segment $A P$, so lines $A P$ and $O T$ are perpendicular. Therefore, lines $B C$ and $O T$ are parallel, and quadrilateral $B O T Q$ is a parallelogram, in particular, $B Q=O T$. By Thales' theorem for parallel lines $C K$ and $O Q$, we have $\frac{B K}{B C}=\frac{B O}{B Q}=\frac{O A}{O T}=\cos \varphi$. Therefore,

$$
B K=B C \cos \varphi=B P \cos \psi \cos \varphi=B E \cos \psi=B E \cos \angle E B K
$$

Thus, point $K$ is the foot of the perpendicular dropped from $E$ to line $B O$, meaning $\angle B K E=90^{\circ}$.