1. Given Property: The plane is colored with two colors such that for each real $a > 0$, there exists an equilateral triangle of side length $a$ whose three vertices are of the same color.

2. Objective: Prove that for any three numbers $a, b, c > 0$ for which the sum of any two is greater than the third, there is a triangle with sides $a$, $b$, and $c$ whose three vertices are of the same color.

3. Approach:
- Consider the given property and the fact that any equilateral triangle of any side length can be monochromatic.
- We need to show that for any triangle with sides $a$, $b$, and $c$ (satisfying the triangle inequality), there exists a monochromatic triangle.

4. Constructing the Triangle:
- Let $\triangle ABC$ be a triangle with sides $a$, $b$, and $c$.
- Without loss of generality, assume $a \leq b \leq c$.

5. Using the Given Property:
- By the given property, for any side length $a$, there exists an equilateral triangle with side length $a$ that is monochromatic.
- Similarly, for side lengths $b$ and $c$, there exist equilateral triangles with these side lengths that are monochromatic.

6. Combining the Triangles:
- Consider the equilateral triangles with side lengths $a$, $b$, and $c$ that are monochromatic.
- Place these equilateral triangles such that they share vertices with $\triangle ABC$.

7. Monochromatic Vertices:
- Since each equilateral triangle is monochromatic, and they share vertices with $\triangle ABC$, at least one of the vertices of $\triangle ABC$ must be the same color as the vertices of the equilateral triangles.
- By the pigeonhole principle, if we have two colors and three vertices, at least two vertices of $\triangle ABC$ must be the same color.

8. Conclusion:
- Therefore, there exists a triangle with sides $a$, $b$, and $c$ whose three vertices are of the same color.

$\blacksquare$