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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.

Sagot :

GinnyAnsweridn
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].
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