PRINCIPIA · THEOREM

Incenter exists — the three angle bisectors of a triangle meet at one point

Dependencies: Angle bisector ⇔ equidistant from sides (on the line ⇒ equidistant from the two sides), Angle bisector test (equidistant from the two sides ⇒ on the bisector).

Statement

Let ABC\triangle ABC be a non-degenerate triangle. Then the bisectors of the three interior angles A\angle A, B\angle B, C\angle C meet at a single point II; this point is called the incenter of ABC\triangle ABC. Equivalently, there is a unique point II inside the triangle such that

dist(I,AB)  =  dist(I,BC)  =  dist(I,CA).\operatorname{dist}(I,\, AB) \;=\; \operatorname{dist}(I,\, BC) \;=\; \operatorname{dist}(I,\, CA).

Denote this common distance by rr; then the circle centered at II with radius rr is tangent to all three sides (the incircle, treated as a separate theorem in the next section).

The incenter I together with the three perpendicular segments of equal length to the three sides

First 20 free · sign in for #21 onward

Sign in to unlock the full proof

The first 20 theorems are free to read; this one and the rest require an account to see the full proof, animation, and consequences. Free, email-code sign-in only.

Sign in to unlock
Help me make this theorem better