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.
Best Answer
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.