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

[Giorgos Giannoulakis]

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

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 [imath]AB>AC[/imath], then how do the areas of triangles [imath]\triangle ABD[/imath] and [imath]\triangle ACD[/imath] compare?

Then, given that comparison, how do [imath]BD[/imath] and [imath]CD[/imath] compare?

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