Euclidean Geometry Problem: Prove that (AB>AC)⟺(BD>CD)

[Giorgos Giannoulakis]

Consider a triangle $$ABC$$ and the bisector $$AD$$ of angle $$A$$. Prove that: $$(AB>AC) \iff (BD>CD)$$
I am struggling with this problem. I am not allowed to use the angle bisector theorem with which of course it is solved very easily.

I was thinking that I need to use the fact that every exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Also the fact that if two angles of a triangle are unequal, then the measures of the sides opposite these angles are also unequal, and the longer side is opposite the greater angle.

Still though I cannot find a way to solve this

Can anyone help me?

Thanks!

 

[Dr.Peterson]

Consider a triangle $ABC$ and the bisector $AD$ of angle $A$. Prove that: $(AB>AC) \iff (BD>CD)$

I am struggling with this problem. I am not allowed to use the angle bisector theorem with which of course it is solved very easily.

I was thinking that I need to use the fact that every exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Also the fact that if two angles of a triangle are unequal, then the measures of the sides opposite these angles are also unequal, and the longer side is opposite the greater angle.

Still though I cannot find a way to solve this

Can anyone help me?

Thanks!

My immediate thought is to use the same approach I would use to prove the angle bisector theorem:

If $AB>AC$, then how do the areas of triangles $\triangle ABD$ and $\triangle ACD$ compare?

Then, given that comparison, how do $BD$ and $CD$ compare?

(This can be written as a sequence of equivalences.)