# bisectors of exterior angles in triangle

In triangle ABC, the bisectors of the exterior angles B and C meet at H. Show that AH is the bisector of the angle BAC.

I was trying to look at some exterior angles in some triangles but I can't find a way to link those two angles that supposed to be equal. thx!

• Do you mean AH, instead of AP? Commented Sep 23, 2022 at 12:21
• yes, thx. I don't know why I was using P :))) Commented Sep 23, 2022 at 12:28

we take the triangle to be made out of lines $$l_1,l_2,l_3$$ with positive direction into the triangle.
the external bisector of $$l_1,l_2$$ is the space of points with directed distance $$d(P,l_1)=-d(P,l_2)$$
and if $$P$$ is on the external bisector of $$l_2,l_3$$ then $$d(P,l_2)=-d(P,l_3)$$
which means $$d(P,l_1)=d(P,l_3)$$ which means it's on the internal bisector of $$l_1,l_3$$.
This is a well-know property of triangle excircles. All points that are equidistant from the sides of external angle $$B$$ lie on its bisector. Ditto for angle $$C$$. The intersection of two bisectors will be a center of excircle which is tangent to the sides of the external angles which are $$BC$$ and extensions of $$AC$$ and $$AB$$. We can see now that from point $$A$$ there are two tangents to the circle thus $$AH$$ will bisect $$\angle BAC$$ (draw two radii to the points of tangency and you have two congruent right triangles). You can read more about excircles and their properties here