4. The diagonals of the inscribed quadrilateral $A B C D$ intersect at point $P$, and triangle $A P D$ is acute-angled. Points $E$ and $F$ are the midpoints of sides $A B$ and $C D$ respectively. A perpendicular is drawn from point $E$ to line $A C$, and a perpendicular is drawn from point $F$ to line $B D$, these perpendiculars intersect at point $Q$. Find the angle between the lines $P Q$ and $B C$.
Answer: $90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-22.jpg?height=1019&width=993&top_left_y=153&top_left_x=540)

First solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $T$ be the point of intersection of the lines $P Q$ and $B C$. Since the angles $\angle B A C$ and $\angle B D C$ subtend the same arc, they are equal. Therefore, the right triangles $A E^{\prime} E$ and $B F^{\prime} F$ are similar. Then, $\frac{E E^{\prime}}{F F^{\prime}}=\frac{A E}{D F}=\frac{B E}{C F}$, the last equality because points $E$ and $F$ are the midpoints of segments $A B$ and $C D$. Moreover, $\angle B E E^{\prime}=\angle C F F^{\prime}$, so triangles $B E E^{\prime}$ and $C F F^{\prime}$ are similar, and in particular, $\angle A B E^{\prime}=\angle D C F^{\prime}$. The angles $\angle A B D$ and $\angle A C D$ subtend the same arc and are therefore equal. Thus, $\angle E^{\prime} B F^{\prime}=\angle A B D-\angle A B E^{\prime}=\angle A C D-\angle D C F^{\prime}=\angle E^{\prime} C F^{\prime}$. Therefore, the quadrilateral $B E^{\prime} F^{\prime} C$ is cyclic and

$$
\angle P B T=\angle C B F^{\prime}=\angle C E^{\prime} F^{\prime}=\angle P E F^{\prime}=\angle P Q F^{\prime}
$$

(the last equality of angles follows from the cyclic nature of the quadrilateral $P E^{\prime} Q F^{\prime}$). It remains to calculate the angles:

$$
\angle P T B=180^{\circ}-\angle P B T-\angle B P T=180^{\circ}-\angle P Q F^{\prime}-\angle Q P F^{\prime}=90^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-23.jpg?height=1128&width=991&top_left_y=156&top_left_x=538)

Second solution. Let $E^{\prime}$ and $F^{\prime}$ be the points of intersection of $E Q$ with $A P$ and $F Q$ with $D P$, respectively, and let $K$ and $L$ be the midpoints of segments $A P$ and $D P$, and $T$ be the point of intersection of the lines $P Q$ and $B C$. Let $O$ be the center of the circumcircle of triangle $A P D$, then $O K$ and $O L$ are the perpendicular bisectors of segments $A P$ and $D P$, respectively. Let $\angle B A P=\varphi$ for brevity. Then $\angle C D P=\varphi$, since the angles $\angle B A P$ and $\angle C D P$ subtend the same arc. Therefore, triangles $B A P$ and $C D P$ are similar by two angles and, hence,

$$
\frac{A P}{D P}=\frac{A B}{C D}=\frac{A E}{D F}=\frac{A E \cos \varphi}{D F \cos \varphi}=\frac{A E^{\prime}}{D F^{\prime}}
$$

Thus,

$$
\frac{P E^{\prime}}{P F^{\prime}}=\frac{A P-A E^{\prime}}{D P-D F^{\prime}}=\frac{A P}{D P}=\frac{P K}{P L}
$$

Let the lines $K O$ and $P Q$ intersect at point $K^{\prime}$, and the lines $L O$ and $P Q$ intersect at point $L^{\prime}$. Then triangles $P E^{\prime} Q$ and $P K K^{\prime}$ are similar with the ratio $\frac{P E^{\prime}}{P K}=\frac{P F^{\prime}}{P L}$, and with the same ratio of similarity, triangles $P F^{\prime} Q$ and $P L L^{\prime}$ are similar. Therefore, $P K^{\prime}=P L^{\prime}$ and, hence, points $K^{\prime}, L^{\prime}$, and $O$ coincide. Thus, the line $P Q$ passes through the point $O$. It remains to calculate the angles:

$\angle P T B=180^{\circ}-\angle C B D-\angle B P T=180^{\circ}-\angle C A D-\angle O P D=180^{\circ}-\angle P O L-\angle O P D=90^{\circ}$.

In the penultimate equality, it was used that the central angle $\angle P O D$ is equal to twice the inscribed angle $\angle C A D$ on one side and to twice the angle $\angle P O L$ on the other side.

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-24.jpg?height=1128&width=985&top_left_y=156&top_left_x=547)

Third solution. Draw a line $\ell$ from point $P$ perpendicular to side $B C$, let $T$ be its point of intersection with $B C$. Draw lines through points $A$ and $D$ perpendicular to diagonals $A C$ and $B D$, respectively. Let $S$ be their point of intersection. Since $\angle P A S=90^{\circ}=\angle P D S$, points $A, P, D$, and $S$ lie on a circle with diameter $P S$. Then (the equality $\angle D A P=\angle D B C$ follows from the cyclic nature of quadrilateral $A B C D$)

$$
\angle D P S=\angle D A S=90^{\circ}-\angle D A P=90^{\circ}-\angle D B C=\angle B P T
$$

Therefore, point $S$ lies on line $\ell$.

Let $H$ be the orthocenter of triangle $P B C$. Clearly, it also lies on line $\ell$. Then the lines $A S, B H$, and $E Q$ are parallel, and since $A E=E B$, line $E Q$ is the midline of trapezoid $A B H S$. Therefore, $E Q$ intersects line $\ell$ at the midpoint of segment $S H$. Similarly, line $F Q$ also intersects line $\ell$ at the midpoint of segment $S H$. But then lines $E Q, F Q$, and $\ell$ intersect at the midpoint of segment $S H$ and this point is point $Q$. Thus, $Q$ also lies on line $\ell$. In particular, lines $P Q$ and $B C$ intersect at a right angle.