Given: A-B-C, D is not on line AC so BD > CD.
Prove: AD > BD
See diagram:
BD > CD Given
$\displaystyle \angle C > \angle CBD$ If one side of a triangle (BD) is longer than another side (CD), then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.
$\displaystyle \angle ABD > \angle C$ If an angle is an exterior angle of a triangle, then its measure is greater than either of its corresponding remote interior angles.
$\displaystyle \therefore AD > BD$ If one angle of a triangle ($\displaystyle \angle ABD$) has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle ($\displaystyle \angle C$).