Problem 1610

Let's break down the problem and solution into detailed steps.

### Part (a): Prove that $R$ lies on $XY$.

1. Given:
- Two circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$.
- Line passing through $P$ intersects $\omega_1$ at $A$ and $\omega_2$ at $B$.
- Another line parallel to $AB$ intersects $\omega_1$ at $D$ and $F$, and $\omega_2$ at $E$ and $C$ such that $E$ and $F$ lie between $C$ and $D$.
- $X$ is the intersection of $AD$ and $BE$.
- $Y$ is the intersection of $BC$ and $AF$.
- $R$ is the reflection of $P$ about $CD$.

2. To Prove:
- $R$ lies on $XY$.

3. Proof:
- We need to show that $\triangle RCY \sim \triangle BXY$ and $\triangle PFC \sim \triangle ABX$.
- Since $AB \parallel CD$, we have $\angle ABX = \angle PCD = \angle DCB$.
- This implies $\angle XBY = \angle RCY$.
- By similarity, $\frac{PC}{CF} = \frac{BX}{AB} = \frac{CR}{CY}$.
- Therefore, $\frac{CR}{BX} = \frac{CY}{BY}$.
- Since $\angle XBY = \angle RCY$, we have $\triangle XBY \sim \triangle CRY$.
- Hence, $\angle BYX = \angle CYX$.

Thus, $R$ lies on $XY$.

### Part (b): Prove that $PR$ is the bisector of $\angle XPY$.

1. Given:
- Same setup as in part (a).

2. To Prove:
- $PR$ is the bisector of $\angle XPY$.

3. Proof:
- We need to show that $\angle XPR = \angle RPY$.
- From part (a), we have $\triangle PFY \sim \triangle PDX$.
- This implies $\frac{PF}{DX} = \frac{CF}{DE} = \frac{CY}{PE} = \frac{FY}{PD}$.
- Therefore, $\triangle PFY \sim \triangle PDX$.
- Hence, $\frac{PY}{PX} = \frac{CF}{DE} = \frac{\frac{CY}{BY}}{\frac{XD}{AX}} = \frac{\frac{RY}{XY}}{\frac{RX}{XY}} = \frac{RY}{RX}$.
- This implies $\angle XPR = \angle RPY$.

Thus, $PR$ is the bisector of $\angle XPY$.