In $\Delta ABC$,$AB = AC$ and $AD$ is an altitude to the base $BC$. Prove that $D$ is the midpoint of $BC$.

(N/A) Given: In $\Delta ABC$,$AB = AC$ and $AD \perp BC$.
To prove: $BD = DC$.
Proof: Consider $\Delta ABD$ and $\Delta ACD$.
$1$. $AB = AC$ (Given).
$2$. $\angle ADB = \angle ADC = 90^{\circ}$ (Since $AD$ is an altitude).
$3$. $AD = AD$ (Common side).
By the $RHS$ (Right angle-Hypotenuse-Side) congruence criterion,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = DC$.
This implies that $D$ is the midpoint of $BC$.