Points on an Angle Bisector

Here is a picture.

In the figure, if $\angle BAD = \angle CAD$ the internal angle $BAC$ is bisected, implying $AD$ is its internal angle bisector.

It is important to note that all points on the internal angle bisector of an angle will be equidistant from the sides of the angle.

For instance, in the figure above, let $E,F$ be the feet of the perpendiculars from $D$ to $AB$ and $AC$.

To prove the claim, we must show $DE = DF$.

Note that, by the definition of an angle bisector, we have $\angle EAD = \angle FAD$ and $DA=DA$. Meanwhile, $\angle DEA = \angle DFA = 90$. This implies that

$\triangle DEA \cong \triangle DFA$ and then it follows that $DE=DF$.

You can show yourself that all points equidistant from $AB,AC$ which lie "inside" interior angle $BAC$ must be on the (internal) angle bisector of $BAC$. Try it!