Let $ABC$ be an acute triangle with side lengths $AB = 7$, $BC = 12$, $AC = 10$, and let $\omega$ be its incircle. If $\omega$ is touching $AB$, $AC$ at $F, E$, respectively, and if $EF$ intersects $BC$ at $X$, suppose that the ratio in which the angle bisector of $\angle BAC$ divides the segment connecting the midpoint of $EX$ and $C$ is $\frac{a}{b}$, where $a, b$ are relatively prime integers. Find $a + b$.

Let's think step by step and output the final answer within \boxed{}.
user