Excenters and External Angle Bisectors

Draw the external angle bisectors at $B$ and $C$, and let their intersection be $I_A$. We prove that $I_A$ also lies on the internal angle bisector at $A$.

Draw perpendiculars from $I_A$ to $BC$ and to the lines $AB$ and $AC$. Denote their lengths by $x$, $z$, and $y$, respectively.

Since $I_A$ lies on the external angle bisector at $B$, its distances to the lines $BA$ and $BC$ are equal. Thus $x=z$.

Since $I_A$ lies on the external angle bisector at $C$, its distances to the lines $CA$ and $CB$ are equal. Thus $x=y$.

Hence $I_A$ is equidistant from the lines $AB$ and $AC$. Because $I_A$ lies inside angle $BAC$, it belongs to the internal angle bisector at $A$. Therefore, the three angle bisectors are concurrent.

Set $x=y=z=r_A$. The circle centered at $I_A$ with radius $r_A$ is tangent to $BC$ and to the extensions of $AB$ and $AC$. This is the $A$-excircle; $I_A$ is its center and $r_A$ is its exradius.

Therefore,

Let $X$ lie on the extension of $AB$ beyond $B$. By the exterior angle theorem,

Since $BI_A$ bisects the exterior angle at $B$,

Since $BX$ is an extension of $BA$, angle $I_ABX$ is an exterior angle of triangle $ABI_A$. Thus,

Similarly,

Since ray $I_AA$ lies between rays $I_AB$ and $I_AC$,

Prove that

Let $I$ be the incenter and $I_A$ the $A$-excenter of triangle $ABC$. Prove that

The three exradii are equal, and the incenter and excenters form a symmetric configuration.

If AB=AC, the A-excenter lies on the symmetry axis of the triangle. The B-excenter and C-excenter are reflections of each other across this axis.

If C=90 degrees, the exradii have particularly simple expressions in terms of the legs and hypotenuse. For example,

where c=AB is the hypotenuse.

The triangle $I_AI_BI_C$ formed by the three excenters is called the excentral triangle.

The original vertices $A$, $B$, and $C$ lie on its respective altitude lines, and the incenter $I$ of $ABC$ is the orthocenter of the excentral triangle.

For the incenter $I$ and the $A$-excenter $I_A$,

whereas

Thus,

Therefore, $B$, $I$, $C$, and $I_A$ are concyclic.

Combining the exradius formula with Heron's formula gives

Similarly,

The inradius satisfies

but the A-exradius satisfies