Sure, let's prove the given inequality by breaking down the steps clearly.
### Given:
We have a triangle [tex]\( \triangle ABC \)[/tex].
### To Prove:
[tex]\[ BC + AC > AB \][/tex]
### Proof:
1. Draw the perpendicular line from vertex [tex]\( C \)[/tex] to segment [tex]\( AB \)[/tex]:
Consider drawing a perpendicular line [tex]\( CE \)[/tex] from vertex [tex]\( C \)[/tex] to the segment [tex]\( AB \)[/tex]. Let [tex]\( E \)[/tex] be the intersection point of [tex]\( CE \)[/tex] with [tex]\( AB \)[/tex].
[tex]\[
\begin{array}{c}
\text{A} \hspace{-0.1em} \bullet \hspace{1em} \text{E} \\
\hspace{1em} \big\downarrow \\
\text{C} \uparrow \hspace{0.1em} \text{B}
\end{array}
\][/tex]
2. Establish the shortest distance properties:
From the shortest distance theorem, we know:
- The line segment [tex]\( BE \)[/tex] is the shortest distance from point [tex]\( B \)[/tex] to line [tex]\( CE \)[/tex].
- The line segment [tex]\( AE \)[/tex] is the shortest distance from point [tex]\( A \)[/tex] to line [tex]\( CE \)[/tex].
Therefore:
[tex]\[ BC > BE \][/tex]
[tex]\[ AC > AE \][/tex]
3. Combine the inequalities:
If we add these inequalities together, we have:
[tex]\[ BC + AC > BE + AE \][/tex]
4. Substitute:
Notice that [tex]\( BE + AE \)[/tex] equals [tex]\( AB \)[/tex] because segment [tex]\( E \)[/tex] is a point on [tex]\( AB \)[/tex] and:
[tex]\[ BE + AE = AB \][/tex]
5. Replace and conclude:
Hence, substituting [tex]\( BE + AE \)[/tex] in the inequality:
[tex]\[ BC + AC > AB \][/tex]
This concludes our proof. Therefore, we have successfully shown that:
[tex]\[ \boxed{BC + AC > AB} \][/tex]
