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ad-work [718]
3 years ago
12

Given: ABC is a triangle. Prove: BC AC > BA Triangle A B C is shown. In triangle ABC, we can draw a perpendicular line segmen

t 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.
SAT
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

CE

AE

segment addition postulate

HOPE IT HELPS :D

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