Problem 1377

To prove that $E$ lies on the line $UK$ if and only if it lies on the line $VH$, we will proceed with the following steps:

1. **Assume $U, E, K$ are collinear:**
- Since $U$ is the circumcenter of $\triangle ABE$, $U$ lies on the perpendicular bisector of $AB$ and $BE$.
- Since $K$ is the orthocenter of $\triangle CDE$, $K$ lies on the altitudes of $\triangle CDE$.

2. **Show that $UK \perp CD$:**
- If $U, E, K$ are collinear, then $UK$ is a straight line passing through $E$.
- Since $U$ is the circumcenter of $\triangle ABE$, $\angle UEB = 90^\circ - \angle A$.
- Since $K$ is the orthocenter of $\triangle CDE$, $\angle DEK = 90^\circ - \angle D$.
- Therefore, $\angle UEB = \angle DEK$ implies $\angle A = \angle D$.

3. **Prove that quadrilateral $ABCD$ is cyclic:**
- Since $\angle A = \angle D$, the opposite angles of quadrilateral $ABCD$ are equal.
- Hence, quadrilateral $ABCD$ is cyclic.

4. **Prove that $H, E, V$ are collinear:**
- Since $ABCD$ is cyclic, the circumcenter $V$ of $\triangle CDE$ lies on the perpendicular bisector of $CD$ and $DE$.
- The orthocenter $H$ of $\triangle ABE$ lies on the altitudes of $\triangle ABE$.
- By similar angle chasing, we can show that $\angle HEV = 90^\circ - \angle C$.
- Therefore, $H, E, V$ are collinear.

5. Conclude the proof:
- We have shown that if $E$ lies on the line $UK$, then $E$ also lies on the line $VH$.
- Conversely, if $E$ lies on the line $VH$, then $E$ also lies on the line $UK$.
- Therefore, $E$ lies on the line $UK$ if and only if it lies on the line $VH$.

$\blacksquare$