Line from Vertex of Triangle to Incenter is Angle Bisector
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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 $\triangle ABC$.
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 $\triangle ABC$.
$\blacksquare$