### In $\triangle ABC$, with $AB = AC$, prove that the altitude from the vertex $A$ bisects the side $BC$.

Step by Step Explanation:
1. We know that $\triangle ABC$ is an isosceles triangle in which $AB = AC$.

Let $AD$ be the altitude from the vertex $A$ on the side $BC$.

As the altitude from a vertex to the opposite side is perpendicular to the opposite side. $$\implies AD\perp BC$$ Let us now represent the above situation with the help of a figure.
2. We need to prove that $BD = DC.$
3. In the right-angled $\triangle ADB$ and $\triangle ADC,$ we have\begin{aligned} & AD = AD && [\text{Common}] \\ & AB = AC && [\text{Given}] \\ \therefore \space & \triangle ADB \cong \triangle ADC && [\text{By RHS criterion}] \end{aligned}
4. As corresponding parts of congruent triangles are equal, we have$$BD = DC$$
5. Thus, in an isosceles triangle, the altitude from the vertex bisects the base.