Given: ABC is a triangle.

Prove: BC + AC > BA

Triangle A B C is shown.

In triangle ABC, we can draw a perpendicular line segment from vertex C to segment AB. The intersection of AB and the perpendicular is called E. We know that BE is the shortest distance from B to
and that
is the shortest distance from A to CE because of the shortest distance theorem. Therefore, BC > BE and AC > AE. Next, add the inequalities: BC + AC > BE + AE. Then, BE + AE = BA because of the
. Therefore, BC + AC > BA by substitution.