Answered

Given: [tex]\( \triangle ABC \)[/tex]

Prove: [tex]\( BC + AC \ \textgreater \ BA \)[/tex]

In [tex]\(\triangle ABC\)[/tex], we can draw a perpendicular line segment from vertex [tex]\(C\)[/tex] to segment [tex]\(AB\)[/tex]. The intersection of [tex]\(AB\)[/tex] and the perpendicular is called [tex]\(E\)[/tex]. We know that [tex]\(BE\)[/tex] is the shortest distance from [tex]\(B\)[/tex] to [tex]\( \square \)[/tex] and that [tex]\(\square\)[/tex] is the shortest distance from [tex]\(A\)[/tex] to [tex]\(CE\)[/tex] because of the shortest distance theorem. Therefore, [tex]\(BC \ \textgreater \ BE\)[/tex] and [tex]\(AC \ \textgreater \ AE\)[/tex].

Next, add the inequalities:
[tex]\[ BC + AC \ \textgreater \ BE + AE \][/tex]

Then, [tex]\( BE + AE = BA \)[/tex] because of the [tex]\(\square\)[/tex] BA by substitution.

Thus, [tex]\( BC + AC \ \textgreater \ BA \)[/tex].



Answer :

GinnyAnswer
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]

Other Questions