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

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.