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Triangle ABC with incenter I, with angle bisectors (red), incircle (blue), and inradii (green)

The incenter of a triangle is the intersection of its (interior) angle bisectors. The incenter is the center of the incircle. Every nondegenerate triangle has a unique incenter.

Proof of Existence

Consider a triangle $ABC$. Let $I$ be the intersection of the respective interior angle bisectors of the angles $BAC$ and $CBA$. We observe that since $I$ lies on an angle bisector of $BAC$, is equidistant from $AB$ and $CA$; likewise, it is equidistant from $BC$ and $AB$; hence it is equidistant from $BC$ and $BC$ and $CA$ and therefore lies on an angle bisector of $ACB$. Since it lies within the triangle $ABC$, this is the interior angle bisector of $ACB$. Since $I$ is equidistant from all three sides of the triangle, it is the incenter.

It should be noted that this proof parallels that for the existence of the circumcenter.

The proofs of existence for the excenters is the same, except that certain angle bisectors are exterior.

Properties of the Incenter

$\bullet$ The incenter of any triangle lies within the orthocentroidal circle.

$\bullet$ The unnormalised areal coordinates of the incenter are $(a,b,c)$

$\bullet$ Let $D$ be a point on the circumcircle of $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then points $B$, $C$, and $I$ lie on a circle centered at $D$.

[asy]defaultpen(fontsize(8)); pair A=(7,10), B=(0,0), C=(10,0), D, I; I=incenter(A,B,C); draw(A--B--C--A);draw(circumcircle(A,B,C));draw(incircle(A,B,C)); D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circumcircle(A,B,C)); draw(A--D);draw(circumcircle(B,C,I)); dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0)); label("D",D,(0,-1));label("I",I,(-1,1));[/asy]