Answer :
Let's prove that in triangle [tex]\(ABC\)[/tex], the sum of the lengths of two sides is greater than the length of the third side. Specifically, we want to show that [tex]\(BC + AC > BA\)[/tex].
### Step-by-Step Solution:
1. Draw Perpendicular: In triangle [tex]\(ABC\)[/tex], draw a perpendicular line segment from vertex [tex]\(C\)[/tex] to segment [tex]\(AB\)[/tex]. Let the intersection point of this perpendicular with [tex]\(AB\)[/tex] be [tex]\(E\)[/tex].
2. Right Triangle Analysis:
- The triangle [tex]\(BEC\)[/tex] is a right triangle with the right angle at [tex]\(E\)[/tex].
- Similarly, triangle [tex]\(AEC\)[/tex] is a right triangle with the right angle at [tex]\(E\)[/tex].
3. Using the Hypotenuse Property:
- In right triangle [tex]\(BEC\)[/tex], the side [tex]\(BC\)[/tex] is the hypotenuse. By the properties of right triangles, the hypotenuse [tex]\(BC\)[/tex] is the longest side, so [tex]\(BC > BE\)[/tex].
- In right triangle [tex]\(AEC\)[/tex], the side [tex]\(AC\)[/tex] is the hypotenuse. Similarly, [tex]\(AC > AE\)[/tex].
4. Add the Inequalities:
- By adding the inequalities from the right triangles, we get:
[tex]\[ BC > BE \quad \text{and} \quad AC > AE \][/tex]
Adding these two inequalities together, we get:
[tex]\[ BC + AC > BE + AE \][/tex]
5. Using the Line Segment Property:
- By the definition of line segments, the line segment [tex]\(AB\)[/tex] can be broken down as:
[tex]\[ BE + AE = BA \][/tex]
6. Substitute and Conclude:
- Now, substituting [tex]\(BA\)[/tex] into our inequality:
[tex]\[ BC + AC > BE + AE \quad \text{and} \quad BE + AE = BA \][/tex]
Thus, we get:
[tex]\[ BC + AC > BA \][/tex]
### Conclusion:
Therefore, we have proven that in triangle [tex]\(ABC\)[/tex], the sum of the lengths of any two sides is greater than the length of the third side, specifically [tex]\(BC + AC > BA\)[/tex], by using the shortest distance theorem and properties of right triangles.
### Step-by-Step Solution:
1. Draw Perpendicular: In triangle [tex]\(ABC\)[/tex], draw a perpendicular line segment from vertex [tex]\(C\)[/tex] to segment [tex]\(AB\)[/tex]. Let the intersection point of this perpendicular with [tex]\(AB\)[/tex] be [tex]\(E\)[/tex].
2. Right Triangle Analysis:
- The triangle [tex]\(BEC\)[/tex] is a right triangle with the right angle at [tex]\(E\)[/tex].
- Similarly, triangle [tex]\(AEC\)[/tex] is a right triangle with the right angle at [tex]\(E\)[/tex].
3. Using the Hypotenuse Property:
- In right triangle [tex]\(BEC\)[/tex], the side [tex]\(BC\)[/tex] is the hypotenuse. By the properties of right triangles, the hypotenuse [tex]\(BC\)[/tex] is the longest side, so [tex]\(BC > BE\)[/tex].
- In right triangle [tex]\(AEC\)[/tex], the side [tex]\(AC\)[/tex] is the hypotenuse. Similarly, [tex]\(AC > AE\)[/tex].
4. Add the Inequalities:
- By adding the inequalities from the right triangles, we get:
[tex]\[ BC > BE \quad \text{and} \quad AC > AE \][/tex]
Adding these two inequalities together, we get:
[tex]\[ BC + AC > BE + AE \][/tex]
5. Using the Line Segment Property:
- By the definition of line segments, the line segment [tex]\(AB\)[/tex] can be broken down as:
[tex]\[ BE + AE = BA \][/tex]
6. Substitute and Conclude:
- Now, substituting [tex]\(BA\)[/tex] into our inequality:
[tex]\[ BC + AC > BE + AE \quad \text{and} \quad BE + AE = BA \][/tex]
Thus, we get:
[tex]\[ BC + AC > BA \][/tex]
### Conclusion:
Therefore, we have proven that in triangle [tex]\(ABC\)[/tex], the sum of the lengths of any two sides is greater than the length of the third side, specifically [tex]\(BC + AC > BA\)[/tex], by using the shortest distance theorem and properties of right triangles.
