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 $C$


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

Proof

 * Euclid-IV-4.png

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:

Hence by Triangle Side-Side-Side Equality:
 * $\triangle EBD = \triangle FBD$

So:
 * $\angle EBD = \angle FBD$

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

By the same reasoning, :
 * $AD$ is the angle bisector of $A$

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

Hence the result.

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 $\triangle ABC$.