Question

Given: ABC is a triangle. Prove: BC + AC > BA 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, BA + AC > BA by substitution.

Answer 1

CE
AE
segment addition postulate 

Answer 2

Answer:

Step-by-step explanation:

Given: ABC is a triangle.

To prove: BC + AC > BA

Proof: 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 AE is the shortest distance from A to CE because of the shortest distance theorem.

Therefore, $BC>BE$ and $AC>AE$.

Now, add the inequalities, we get

$BC+AC>BE+AE$.

Then, $BE+AE=BA$ because Segment addition postulate (states that given 2 points E and F, a third point D lies on the line segment EF if and only if the distances between the points satisfy the equation ED + DF = EF)

Therefore, $BA+AC>BA$ by substitution.