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].
