Given: $\triangle ABC$ 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 $E$ and that $AE$ is the shortest distance from $A$ to $E$ because of the shortest distance theorem. Therefore, $BC > BE$ and $AC > AE$.

Next, add the inequalities: [tex]$BC + AC \ \textgreater \ BE + AE$[/tex]. Then, [tex]$BE + AE = BA$[/tex] because of the segment addition postulate. Therefore, [tex]$BC + AC \ \textgreater \ BA$[/tex] by substitution.

Certainly! Let's go through the proof step-by-step:

1. Given:
- [tex]$ABC$[/tex] is a triangle.

2. To Prove:
- $ BC + AC > BA $

3. Construction:
- Draw a perpendicular line segment from vertex $ C $ to segment $ AB $. Let the intersection point of $ AB $ and the perpendicular be $ E $.

4. Key Observations:
1. The line segment $ BE $ is the perpendicular distance from $ B $ to $ CE $. By the Shortest Distance Theorem, this perpendicular distance $ BE $ is the shortest distance from $ B $ to $ CE $.
2. Similarly, the line segment $ AE $ is the perpendicular distance from $ A $ to $ CE $, and is the shortest distance from $ A $ to $ CE $.

5. Inequalities:
- Since $ BC $ is a slant distance and $ BE $ is the perpendicular (shortest distance), we have:
[tex]\[ BC > BE \][/tex]
- Similarly, since $ AC $ is a slant distance and $ AE $ is the perpendicular (shortest distance), we have:
[tex]\[ AC > AE \][/tex]

6. Adding Inequalities:
- We can add the inequalities $ BC > BE $ and $ AC > AE $:
[tex]\[ BC + AC > BE + AE \][/tex]

7. Relationship of $ BE $ and $ AE $ to $ BA $:
- Since $ E $ is the intersection point of the perpendicular from $ C $ to $ AB $, and $ BE + AE $ forms a part of the line segment $ AB $, thus:
[tex]\[ BE + AE = BA \][/tex]

8. Substitution:
- Substituting $ BA $ for $ BE + AE $ in the inequality $ BC + AC > BE + AE $, we get:
[tex]\[ BC + AC > BA \][/tex]

Conclusion:
- Therefore, by the Shortest Distance Theorem and the triangle inequality, we have:
[tex]\[ BC + AC > BA \][/tex]

Thus, we have successfully proven that in triangle [tex]$ ABC $[/tex], [tex]$ BC + AC > BA $[/tex].