Question 1209061
**1. Verify Equilateral Triangle**

* **Points:**
    * A: (0, 0)
    * B: (a, 0)
    * C: (c, d) = (a/2, (a√3)/2)

* **Distances:**
    * AB = √[(a - 0)² + (0 - 0)²] = √a² = a
    * BC = √[(a/2 - a)² + ((a√3)/2 - 0)²] = √[(-a/2)² + (a√3/2)²] = √(a²/4 + 3a²/4) = √(4a²/4) = a
    * AC = √[(a/2 - 0)² + ((a√3)/2 - 0)²] = √[(a/2)² + (a√3/2)²] = √(a²/4 + 3a²/4) = √(4a²/4) = a

Since AB = BC = AC = a, the points A, B, and C form an equilateral triangle.

**2. Find Midpoints**

* **Midpoint of AB (M1):** 
    * [(0 + a)/2, (0 + 0)/2] = (a/2, 0)

* **Midpoint of BC (M2):** 
    * [(a + a/2)/2, (0 + (a√3)/2)/2] = (3a/4, (a√3)/4)

* **Midpoint of AC (M3):** 
    * [(0 + a/2)/2, (0 + (a√3)/2)/2] = (a/4, (a√3)/4)

**3. Verify Equilateral Triangle for Midpoints**

* **M1M2:** 
    * √[((3a/4) - (a/2))² + ((a√3)/4 - 0)²] = √[(a/4)² + (a√3/4)²] = √(a²/16 + 3a²/16) = √(4a²/16) = a/2

* **M2M3:** 
    * √[((a/4) - (3a/4))² + ((a√3)/4 - (a√3)/4)²] = √[(-a/2)² + 0²] = a/2

* **M3M1:** 
    * √[((a/2) - (a/4))² + (0 - (a√3)/4)²] = √[(a/4)² + (a√3/4)²] = √(a²/16 + 3a²/16) = a/2

Since M1M2 = M2M3 = M3M1 = a/2, the midpoints M1, M2, and M3 also form an equilateral triangle.

**Therefore, it is proven that if the points (0, 0), (a, 0), and (c, d) (where c = a/2 and d = (a√3)/2) are the vertices of an equilateral triangle, then the midpoints of the sides of this triangle also form an equilateral triangle.**