Problem 1511

1. Define Points and Midpoints:
Let $U$ and $V$ be the midpoints of $BC$ and $AC$ respectively. Let $S$ and $T$ be the midpoints of $PM$ and $QN$ respectively. We need to show that the line passing through $S$ and $T$ is parallel to $AB$.

2. Angle Bisectors and Altitude Intersection:
Since $AM$ and $BN$ are angle bisectors, they intersect the opposite sides at points $M$ and $N$ respectively. The altitude $CH$ intersects $AM$ at $P$ and $BN$ at $Q$.

3. Isosceles Triangle Property:
Notice that $\angle CMP = \angle CPM = 90^\circ - \frac{\angle A}{2}$. This implies that $\triangle CPM$ is isosceles with $CP = CM$. Since $S$ is the midpoint of $PM$, $CS \perp PM$.

4. Cyclic Quadrilateral:
Since $CS \perp PM$ and $S$ is the midpoint of $PM$, quadrilateral $AHSC$ is cyclic with center $V$ (the midpoint of $AC$). This is because $V$ is equidistant from $A$ and $C$.

5. Parallel Lines:
In the cyclic quadrilateral $AHSC$, we have $\angle VSA = \angle VAS = \angle SAH$. This implies that $VS \parallel AB$. Therefore, $S$ lies on the line $UV$.

6. **Symmetry for $T$:**
Similarly, by considering the properties of $\triangle BQN$, we can show that $T$ lies on the line $UV$.

7. Conclusion:
Since both $S$ and $T$ lie on the line $UV$ and $UV \parallel AB$, the line passing through $S$ and $T$ is parallel to $AB$.

$\blacksquare$