PRINCIPIA · THEOREM

Three excircles (excenters) exist — one interior bisector + two exterior bisectors are concurrent

Dependencies: Angle bisector ⇔ equidistant from sides (including its converse: a point equidistant from the two sides of an angle lies on its bisector), External angle bisector property (a point on the exterior bisector is equidistant from the opposite side and the extension of an adjacent side).

Statement

Let ABC\triangle ABC be a non-degenerate triangle. Fix a vertex (say AA); then the interior bisector of A\angle A and the exterior bisectors of B\angle B and C\angle C all meet at a single point IAI_A. This point is called the excenter of ABC\triangle ABC opposite vertex AA. Equivalently, IAI_A is at equal distance from the three lines BCBC, ABAB, ACAC:

dist(IA,BC)  =  dist(IA,AB)  =  dist(IA,AC).\operatorname{dist}(I_A,\, BC) \;=\; \operatorname{dist}(I_A,\, AB) \;=\; \operatorname{dist}(I_A,\, AC).

Denote this common distance by rAr_A; then the circle centered at IAI_A with radius rAr_A is simultaneously tangent to the interior of side BCBC, to the extension of ABAB beyond BB, and to the extension of ACAC beyond CC — this is the excircle ωA\omega_A of ABC\triangle ABC. Each of the three vertices yields an excenter, so there are three excenters IAI_A, IBI_B, ICI_C and three excircles.

\triangle ABC + excenter I_A + excircle \omega_A (dashed), tangent to BC and to the extensions of AB and AC

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