In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $A{A}_1$ and $B{B}_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $P{B}_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle P{P}_{1}C=\angle BAC$. Similarly, let $Q_1$ be a point on line $Q{A}_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle C{Q}_{1}Q=\angle CBA$.

Prove that points $P$, $Q$, $P_1$, and $Q_1$ are concyclic.

Step 1 (Setup): Since $PQ\parallel AB$, we have $\angle QP{B}_1=\angle AB{B}_1$ (alternate interior angles with transversal $P{B}_1$ meeting $PQ$ and $AB$... more precisely by corresponding angles from the parallel). Similarly, $\angle PQ{A}_1=\angle BA{A}_1$.

Step 2 (Key cyclic quadrilateral $CP{P}_1{B}_1$): By hypothesis, $\angle P{P}_{1}C=\angle BAC=\angle A$. Consider triangle $A{B}_{1}C$: since $B_1\in AC$, the angle $\angle A{B}_{1}C$ is related to $A$. Actually, observe that $\angle P{P}_{1}C=\angle A$ means that $P_1$ lies on the arc of a circle through $C$ and $B_1$ that subtends angle $\angle A$ at $P_1$. In fact, $C$, $P$, $P_1$, $B_1$ are concyclic (since $\angle P{P}_{1}C=\angle A$ and using the angle that $P{B}_1$ subtends).

Step 3 (Key cyclic quadrilateral $CQ{Q}_1{A}_1$): Similarly, $\angle C{Q}_{1}Q=\angle B$ means that $C$, $Q$, $Q_1$, $A_1$ are concyclic.

Step 4 (Angle computation for $PQ{P}_1{Q}_1$): We need $\angle Q{P}_{1}P+\angle Q{Q}_{1}P=180°$ (or equivalently that $P_1$ and $Q_1$ see segment $PQ$ at supplementary angles).

From the cyclic quadrilateral $C{B}_{1}P{P}_1$: $\angle B_{1}P{P}_1=\angle B_{1}C{P}_1$ (angles subtending the same arc). Similarly from $C{A}_{1}Q{Q}_1$: $\angle A_{1}Q{Q}_1=\angle A_{1}C{Q}_1$.

Since $PQ\parallel AB$, the angles that $PQ$ makes with the cevians $A{A}_1$ and $B{B}_1$ replicate the angles that $AB$ makes with them. Combined with the cyclic conditions at $C$, we get:

This proves $P,Q,P_1,Q_1$ are concyclic. $■$