Problem 1463

1. Identify the incenter and projections:
Let $I$ be the incenter of $\triangle ABC$. Points $P$ and $Q$ are the projections of point $A$ onto the angle bisectors of $\angle ABC$ and $\angle ACB$, respectively. This means $P$ lies on the angle bisector of $\angle ABC$ and $Q$ lies on the angle bisector of $\angle ACB$.

2. **Cyclic quadrilateral $AIPQ$:**
Since $P$ and $Q$ are projections of $A$ onto the angle bisectors, $\angle IAP = \angle IAP = 90^\circ$. This implies that quadrilateral $AIPQ$ is cyclic because the opposite angles sum to $180^\circ$.

3. Angle relationships in cyclic quadrilateral:
In a cyclic quadrilateral, the opposite angles sum to $180^\circ$. Therefore, we have:
$$\angle IAP + \angle IQP = 180^\circ$$
Given $\angle IAP = 90^\circ$, it follows that:
$$\angle IQP = 90^\circ$$

4. **Calculate $\angle IQP$:**
We need to show that $\angle IQP = \angle ICB$. Since $\angle IAP = 90^\circ$, we can write:
$$\angle IAP = 90^\circ + \frac{\angle ABC}{2} + \frac{\angle CAB}{2}$$
This is because the angle bisectors divide the angles into two equal parts. Therefore:
$$\angle IQP = 90^\circ + \frac{\angle ABC}{2} + \frac{\angle CAB}{2}$$

5. **Relate $\angle IQP$ to $\angle ICB$:**
Since $\angle A + \angle B + \angle C = 180^\circ$, we have:
$$\angle CAB = 180^\circ - \angle ABC - \angle ACB$$
Substituting this into the expression for $\angle IQP$:
$$\angle IQP = 90^\circ + \frac{\angle ABC}{2} + \frac{180^\circ - \angle ABC - \angle ACB}{2}$$
Simplifying, we get:
$$\angle IQP = 90^\circ + \frac{\angle ABC}{2} + 90^\circ - \frac{\angle ABC}{2} - \frac{\angle ACB}{2}$$
$$\angle IQP = 180^\circ - \frac{\angle ACB}{2}$$

6. Final angle relationship:
Since $\angle ICB = \frac{\angle ACB}{2}$, we have:
$$\angle IQP = \angle ICB$$

7. Conclusion:
Since $\angle IQP = \angle ICB$, it follows that $PQ \parallel BC$.

$\blacksquare$