Prove that AM is perpendicular to BC.

From the problem statement, we are given that $AB = AC$.
From the figure, we can observe:

• $AB = AC$ (indicated by the tick marks on sides $AB$ and $AC$).
• $BM = MC$ (indicated by the tick marks on segments $BM$ and $MC$).
• $AM$ is a common side to both $\triangle AMB$ and $\triangle AMC$.

Consider $\triangle AMB$ and $\triangle AMC$:

1. $AB = AC$ (Given)
2. $BM = MC$ (Given from the figure)
3. $AM = AM$ (Common side)

By the Side-Side-Side (SSS) congruence criterion, $\triangle AMB \cong \triangle AMC$.

Since $\triangle AMB \cong \triangle AMC$, their corresponding angles are equal.
Therefore, $\angle AMB = \angle AMC$.

Angles $\angle AMB$ and $\angle AMC$ form a linear pair on the straight line $BC$.
The sum of angles in a linear pair is $180^\circ$.

Since $\angle AMB = \angle AMC$, we can substitute $\angle AMB$ for $\angle AMC$:

Since $\angle AMB = \angle AMC$, it follows that $\angle AMC = 90^\circ$.

Therefore, $\angle BMA = \angle CMA = 90^\circ$.

$\text{Proven that } \angle BMA = \angle CMA = 90^\circ$