# Thread: triangle edge and distance of vertice to bisector intersection

1. ## triangle edge and distance of vertice to bisector intersection

I am not sure how to geometrically prove: In a triangle ABC, if the bisector of angle A intersects the edge BC at a point D, then AB > BD and AC > CD.

2. ## Re: triangle edge and distance of vertice to bisector intersection

$\frac{\text{BD}}{\text{AB}}=\frac{\text{CD}}{\text {AC}}=\frac{\text{BD}+\text{CD}}{\text{AB}+\text{A C}}=\frac{\text{BC}}{\text{AB}+\text{AC}}<1$

Thanks.