Problem 1057

## 1st solution

We denote the two given angular sizes with $\alpha=\angle A B C=\angle C Q D$ and $\beta=\angle B C D=\angle A P B$. Because $A B \| C D$ applies $\alpha+\beta=180^{\circ}$. Also because $A B \| C D$, it follows from $A K / K B=D L / L C$ that the lines $A D, B C$ and $K L$ intersect at a common point $S$. We consider the centric stretching at $S$, which maps the point $D$ to $A$ ($L$ is mapped to $K$ and $C$ to $B$). The image point of $Q$ in this mapping is $Z$. The quadrilateral $A Z B P$ is a chordal quadrilateral, because

$$\angle A Z B+\angle A P B=\angle D Q C+\angle A P B=\alpha+\beta=180^{\circ}$$

Let $x=\angle S Q C=\angle S Z B$. From the peripheral angle theorem in the chordal quadrilateral $A Z B P$ follows $\angle P A B=x$. Using the sum of angles in the triangle $A B P$, we obtain $\angle A B P=\alpha-x$ and therefore

$$\angle P B C=\alpha-\angle A B P=\alpha-(\alpha-x)=x$$

The following therefore applies $\angle S Q C=\angle P B C$ and from this it follows - regardless of the order in which $P$ and $Q$ lie on the line $K L$ - that the points $P, Q, B$ and $C$ lie on a circle.