The problem statement tells us that
Base angles of an isosceles triangle are equal, so ∠MAB ≅ ∠MBA. ∠AMC is the exterior angle opposite those two, so it is equal to their sum, 2∠MAB.
The base angles in isosceles ΔAMC are equal to half the difference between the apex angle, ∠AMC, and 180°. That is,
... ∠MAC = (1/2)(180° -∠AMC) = (1/2)(180° -2∠MAB) = 90° -∠MAB
The angle at A of ΔABC is the sum of the two angles created by the median AM. That is ...
∠A = ∠MAC + ∠MAB
∠A = (90° -∠MAB) +∠MAB
∠A = 90°
_____
Maybe a shorter way to get there is to realize that ...
... MA ≅ MB ≅ MC
so M is the center of a circle with BC as a diameter and A a point on the circle. Angle BAC is inscribed in the semicircle and subtends an arc of 180°, so angle A is 90°.
