Problem 1373

1. Let $O_1$ and $O_2$ be the centers of circles $C_1$ and $C_2$, respectively. Since the circles touch externally at point $P$, the line $O_1O_2$ passes through $P$.

2. Let $Q$ be a point on $C_1$ such that the tangent at $Q$ intersects $C_2$ at points $A$ and $B$. The line $QP$ intersects $C_2$ again at point $C$.

3. Since $Q$ is a point on $C_1$ and $PQ$ is a tangent to $C_1$ at $Q$, we have $O_1Q \perp PQ$.

4. Let $\alpha = \angle PQA$. Since $PQ$ is tangent to $C_1$ at $Q$, we have $\angle PQO_1 = 90^\circ - \alpha$.

5. Since $P$ is the point of tangency and the circles touch externally, $\angle QPO_1 = \alpha$.

6. Similarly, since $C$ lies on $C_2$ and $QP$ intersects $C_2$ at $C$, we have $\angle PCO_2 = \alpha$.

7. Since $O_1O_2$ is the line joining the centers of the two circles and passes through $P$, we have $\angle PO_2C = 90^\circ$.

8. Therefore, $O_1O_2 \parallel QB$. This implies that $CO_2 \perp AB$.

9. Since $CO_2 \perp AB$, $CO_2$ is the perpendicular bisector of $AB$. The perpendicular bisector of any chord of a circle passes through the center of the circle.

10. Let $CO_2 \cap AB = K$. By the SSS (Side-Side-Side) congruence criterion, $\triangle AO_2K \cong \triangle BO_2K$.

11. Hence, $AK = BK$, which means $K$ is the midpoint of $AB$.

12. Since $CO_2$ is the perpendicular bisector of $AB$, $\triangle ABC$ is isosceles with $AC = BC$.

$\blacksquare$