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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.

Sagot :

Answer:

CE

AE

segment addition postulate

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Answer:

CE

AE

segment addition postulate

Explanation:

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 CE and that AE  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 segment addition postulate . Therefore, BC + AC > BA by substitution.

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