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.