, 2019

Solution 1. Throughout the solution we use oriented angles.
Let rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By
$$\angle Q P A_{2}=\angle B A A_{2}=\angle B B_{2} A_{2}=\angle Q B_{2} A_{2},$$
points $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\omega$.

By
$$\angle C A_{2} A_{1}=\angle C A_{2} A=\angle C B A=\angle C Q_{1} Q=\angle C Q_{1} A_{1},$$
points $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get
$$\angle Q Q_{1} A_{2}=\angle A_{1} Q_{1} A_{2}=\angle A_{1} C A_{2}=\angle B C A_{2}=\angle B A A_{2}=\angle Q P A_{2},$$
so $Q_{1}$ lies on $\omega$.
It follows similarly that $P_{1}$ lies on $\omega$.

Solution 2. First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$.
Let line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then
$$\angle P P_{1} C=\angle B A C=\angle P E C,$$
so points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\omega_{Q}$.
Let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \cap B A_{1}, R=A_{1} Q \cap B_{1} P$ and $T=A Q \cap B P$ are collinear.
Let line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \| P Q$ we obtain
$$\frac{S P}{S Q}=\frac{U B}{U A}=\frac{S F}{S E},$$
so
$$S P \cdot S E=S Q \cdot S F .$$

So, point $S$ has equal powers with respect to $\omega_{P}$ and $\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \cdot R P_{1}=R Q \cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic.
Now consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$.
Let line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \cdot S E=S Q \cdot S F$, so $S$ lies on the radical axis between $\omega_{P}$ and $\omega_{Q}$.

Line $C S T$, that is the radical axis between $\omega_{P}$ and $\omega_{Q}$, is perpendicular to the line $\ell$ of centres of $\omega_{P}$ and $\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\ell$, and hence is cyclic.