1. Consider the triangle $ABC$ with the incircle tangent to sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Let $AD$ intersect the incircle again at point $Q$.

2. We need to show that the line $EQ$ meets the segment $AF$ at its midpoint if and only if $AC = BC$.

3. Let $M$ be the midpoint of $AF$. We need to prove that $M$ lies on $EQ$ if and only if $AC = BC$.

4. First, assume $M$ lies on $EQ$. This implies that $AM = MF$.

5. By the Power of a Point theorem, we have:
$$AM^2 = MQ \cdot ME$$

6. This implies that $\triangle MQA \sim \triangle MAE$ by the similarity criterion (since $AM$ is common and the angles are equal).

7. Therefore, $\angle MAQ = \angle AEM$.

8. Since $M$ is the midpoint of $AF$, we have $\angle FAD = \angle AEQ$.

9. Considering the angles, we have:
$$\angle BFD - \angle QDF = \angle AEF - \angle QEF$$

10. Simplifying the angles, we get:
$$\alpha + \gamma - \angle QDF = \beta + \gamma - \angle QDF$$

11. This simplifies to:
$$2\alpha = 2\beta$$

12. Hence, $\alpha = \beta$, which implies $\angle BAC = \angle CBA$.

13. Therefore, $AC = BC$.

14. Conversely, assume $AC = BC$. Then $\angle BAC = \angle CBA$.

15. This implies that the triangle $ABC$ is isosceles with $AC = BC$.

16. Since $AC = BC$, the incircle is symmetric with respect to the angle bisector of $\angle BAC$.

17. Therefore, the line $EQ$ will meet $AF$ at its midpoint $M$.

18. Hence, $M$ lies on $EQ$ if and only if $AC = BC$.

$\blacksquare$