# How to show angle bisector in triangle ABC?

I was giving the following question:

Let ABC be a triangle. The outer angle bisector of B and the outer angle bisector of C meet in point O. I need to show that AO is the angle bisector of A.

In the question there's some guidance that says that from point O we should draw a perpendicular line to BC and to AB and to AC.

I've been trying to solve this question for a couple of days now and still no success. I think that if I can show angle CBO is equal to BCO then that will allow me to solve the question.

I should also mention that I've basic understanding of geometry so I assume the answer is probably using mostly congruence sentences etc.

Thanks for your help.

-
Angles CBO and BCO will not be equal unless ABC and ACB are equal. Also, ABOC will not normally be a rectangle. – AMPerrine Oct 28 '11 at 16:31
@AMPerrine I added an image to show the drawing that was given with the question. Do you have an idea of how to solve this question? – Einat Oct 28 '11 at 17:11

Compare the distances of $O$ to the three sides of the original triangles.

Having $O$ on an angle bisector means exactly that some of these distances are the same. This may be a theorem that you already know or you have to prove it using congruence with the help of the equal half-angles.

Therefore, you can conclude $dist(O,BC)=dist(O,AB)$ and $dist(O,CA)=dist(O,CB)$ which gives $dist(O,AB) = dist(O,AC)$.

So $O$ also lies on an angle bisector of $A$ and since it lies inside the angle at $A$ of the triangle, this has to be the internal bisector.

-

by incentre excentre configuration,

incentre I , excentre O , vertex A are colinear,

therefore , by joining AO , incentre I also lies on AO.

thus AO bisects angle BAC.

-