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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!

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

2 Answers 2

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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 $l_1,l_3$.

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  • $\begingroup$ well, nice one. but I want to solve it using just euclidean geometry. I am sorry for my english $\endgroup$
    – Numbers
    Commented Sep 23, 2022 at 13:39
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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

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