1. **Problem Statement:** Given an equilateral triangle $ABC$ with side length $2\sqrt{a}$, a point $P$ inside the triangle satisfies $PA=2$, $PB=2\sqrt{3}$, and $PC=4$. We need to find the value of $a$. 2. **Key Formula (British Flag Theorem):** For any point $P$ inside a rectangle, the sum of the squares of the distances to two opposite vertices equals the sum of the squares of the distances to the other two opposite vertices. However, for an equilateral triangle, we use the **British Flag Theorem** adapted for triangles: $$PA^2 + PB^2 + PC^2 = AB^2 + BC^2 + CA^2$$ Since $ABC$ is equilateral, $AB=BC=CA=2\sqrt{a}$. 3. **Apply the formula:** Calculate the sum of squares of distances from $P$: $$PA^2 = 2^2 = 4$$ $$PB^2 = (2\sqrt{3})^2 = 4 \times 3 = 12$$ $$PC^2 = 4^2 = 16$$ Sum: $$4 + 12 + 16 = 32$$ 4. **Sum of squares of sides:** Each side length squared: $$(2\sqrt{a})^2 = 4a$$ Sum of squares of all three sides: $$3 \times 4a = 12a$$ 5. **Equate and solve for $a$:** $$PA^2 + PB^2 + PC^2 = AB^2 + BC^2 + CA^2$$ $$32 = 12a$$ Divide both sides by 12: $$a = \frac{32}{12} = \frac{8}{3}$$ **Final answer:** $$\boxed{\frac{8}{3}}$$