Question 155
Let $\triangle ABC$ be an acute triangle. The lines $(\varepsilon_{1}),(\varepsilon_{2})$ are perpendicular to $AB$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the sides of the triangle $AC; BC$ intersect the lines $(\varepsilon_{1}),(\varepsilon_{2})$ at the points $E, F$, respectively. If $I$ is the intersection point of $EF, MC$, prove that
$$
\angle AIB = \angle EMF = \angle CAB + \angle CBA
$$
$$
\angle AIB = \angle EMF = \angle CAB + \angle CBA
$$