Problem 1357

Given:
- $C_1$ is a circle.
- $P$ is a fixed point outside the circle $C_1$.
- Quadrilateral $ABCD$ lies on the circle $C_1$ such that rays $AB$ and $CD$ intersect at $P$.
- $E$ is the intersection of $AC$ and $BD$.

We need to prove:
(a) The circumcircle of triangle $ADE$ and the circumcircle of triangle $BEC$ pass through a fixed point.
(b) Find the locus of point $E$.

### Part (a)
1. Construct Tangents and Points:
- Let $PK$ and $PQ$ be two tangents from $P$ to $C_1$.
- Let $O$ be the center of $C_1$.
- Let $H$ be the intersection of $PO$ and $KQ$.
- Let $CH$ intersect $C_1$ at $J$.
- Let $BH$ intersect $C_1$ at $F$.

2. Cyclic Quadrilateral:
- Since $HA \cdot HO = HK \cdot HQ = HC \cdot HJ$, it follows that $PJOC$ is cyclic.
- Therefore, $\angle OJH = \angle OPC$.

3. Similar Triangles:
- Since $OH \cdot OP = R^2 = OD^2$, we have $\frac{OH}{OD} = \frac{OD}{OA}$.
- Thus, $\triangle HOD \sim \triangle DOP$, implying $\angle OPC = \angle ODH$.

4. Perpendicularity and Parallelism:
- From the above, $\angle HJO = \angle HDO$, so $HO \perp JD$.
- Therefore, $JD \parallel KQ$.
- Similarly, $AF \parallel KQ \parallel JD$, implying $\text{arc } AD = \text{arc } JF$.

5. Cyclic Quadrilaterals:
- We have $\angle BHC = \angle BEC$, so $BHEC$ is cyclic.
- Similarly, $AHED$ is cyclic.

6. Conclusion:
- The circumcircles of triangles $ADE$ and $BEC$ pass through a fixed point $H$.

### Part (b)
1. Angle Calculation:
- We have $180^\circ - \angle BHE = \angle BCE = \frac{1}{2} (\text{arc } BK + \text{arc } KA) = \frac{1}{2} (\text{arc } BK + \text{arc } QF) = \angle BHK$.

2. Collinearity:
- Therefore, $K, H, E$ are collinear.

3. **Locus of $E$:**
- Hence, $E \in [KQ]$.

The final answer is $\boxed{ E \in [KQ] }$.