Problem 10.7. Given a right triangle $ABC$ with legs $AB=42$ and $BC=56$. A circle passing through point $B$ intersects side $AB$ at point $P$, side $BC$ at point $Q$, and side $AC$ at points $K$ and $L$. It is known that $PK=KQ$ and $QL: PL=3: 4$. Find $PQ^2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-13.jpg?height=502&width=442&top_left_y=92&top_left_x=498)
Answer: 1250.

Solution. Since in a cyclic quadrilateral the sum of opposite angles is $180^{\circ}$, then $\angle P K L=\angle P L Q=90^{\circ}$. From the condition, it also follows that right triangles $A B C$ and $Q L P$ are similar (Fig. 6). From this similarity and the cyclic nature of the pentagon

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-13.jpg?height=502&width=426&top_left_y=839&top_left_x=510)

Fig. 6: to the solution of problem 10.7

$B Q L K P$, we get that $\angle C=\angle Q P L=\angle Q K L$, so triangle $C Q K$ is isosceles and $C Q=Q K$. Similarly, $\angle A=\angle P Q L=180^{\circ}-\angle P K L=\angle A K P$, so $A P=P K$. Also, by the condition, $P K=K Q$.

Thus, $C Q=Q K=P K=A P=x$. From the condition, we have

$$
A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{14^{2}\left(3^{2}+4^{2}\right)}=14 \cdot 5=70
$$

and also $\cos A=\frac{3}{5}$ and $\cos C=\frac{4}{5}$. Drop perpendiculars $P P^{\prime}$ and $Q Q^{\prime}$ to the hypotenuse $A C$. Then

$$
70=A C=A K+K C=2 A P^{\prime}+2 C Q^{\prime}=2 x \cos A+2 x \cos C=2 x\left(\frac{3}{5}+\frac{4}{5}\right)=\frac{14}{5} x
$$

from which $x=25$. By the Pythagorean theorem in triangle $P K Q$, we get

$$
P Q^{2}=P K^{2}+Q K^{2}=25^{2}+25^{2}=1250
$$

Another solution. As in the previous solution, by the Pythagorean theorem, we find $A C=70$. Also, from the similarity and the cyclic nature, we get $\angle C=\angle Q P L=\angle Q B L$, so triangle $C L B$ is isosceles and $C L=L B$. Similarly, $\angle A=\angle P Q L=\angle P B L$, so $A L=B L=C L$, i.e., $L$ is the midpoint of the hypotenuse.

Point $K$ is equidistant from points $P$ and $Q$, i.e., it is the midpoint of the arc $P L Q$ of the given circle and lies on the angle bisector of $\angle B$. Then, by the property of the angle bisector, we get $A K: K C=$ $A B: B C=3: 4$, which gives us $A K=\frac{3}{7} A C=30$ and $C K=\frac{4}{7} A C=40$. From the properties of secants to a circle drawn from the same point, we get $A K \cdot A L=A B \cdot A P$ and $C K \cdot C L=C Q \cdot C B$, from which respectively

$$
A P=\frac{A K \cdot A L}{A B}=\frac{30 \cdot 35}{42}=25 \quad \text { and } \quad C Q=\frac{C K \cdot C L}{C B}=\frac{40 \cdot 35}{56}=25
$$

Then $B P=A B-A P=17$ and $B Q=B C-C Q=31$, from which by the Pythagorean theorem in triangle $P B Q$ we get $P Q^{2}=P B^{2}+B Q^{2}=17^{2}+31^{2}=1250$.