Line from Vertex of Triangle to Incenter is Angle Bisector

Theorem

Let $\triangle ABC$ be a triangle.

Let $D$ be a point in the interior of $\triangle ABC$.

Then:

$AD$ is the angle bisector of $A$

$BD$ is the angle bisector of $B$

$CD$ is the angle bisector of $C$

if and only if:

$D$ is the incenter of $\triangle ABC$.

Proof

Necessary Condition

Let $D$ be the incenter of $\triangle ABC$.

Let $E$, $F$ and $G$ be the points on $AB$, $BC$ and $AC$ tangent to the incircle $\bigcirc EFG$ of $\triangle ABC$.

We have that:

$ED = DF$, as both equal the radius of $\bigcirc EFG$

$\angle BED = \angle BFD$, a right angle

$BD$ is common.

From Pythagoras's Theorem we have that:

\(\ds BD^2\)

\(=\)

\(\ds DE^2 + EB^2\)

Pythagoras's Theorem

\(\ds \)

\(=\)

\(\ds DF^2 + FB^2\)

Pythagoras's Theorem

\(\ds \leadsto \ \ \)

\(\ds EB\)

\(=\)

\(\ds FB\)

algebra

Hence by Triangle Side-Side-Side Congruence:

$\triangle EBD = \triangle FBD$

So:

$\angle EBD = \angle FBD$

and it is seen that $BD$ is the angle bisector of $B$.

By the same reasoning, mutatis mutandis:

$AD$ is the angle bisector of $A$

and

$CD$ is the angle bisector of $C$.

Hence the result.

$\Box$

Sufficient Condition

Let:

$AD$ be the angle bisector of $A$

$BD$ be the angle bisector of $B$

$CD$ be the angle bisector of $C$

It follows from Inscribing Circle in Triangle that $D$ is the incenter of $\triangle ABC$.

$\blacksquare$