Given: A-B-C, D is not on line AC so BD > CD.

Prove: AD > BD

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- Sep 18th 2008, 11:29 AMGoldendoodleMomProof
Given: A-B-C, D is not on line AC so BD > CD.

Prove: AD > BD - Sep 18th 2008, 12:04 PMmasters
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 ($\displaystyle \angle C$).