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.
Since CE is perpendicular to AB at E, BE is the shortest distance from point B to segment CE. Likewise, AE is the shortest distance from point A to CE. Thus the missing segment is CE.
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