Prove that $(2a, 4a)$, $(2a, 6a)$, and $(2a + \sqrt{3}a, 5a)$ are the vertices of an equilateral triangle.

(N/A) Let the vertices be $A(2a, 4a)$, $B(2a, 6a)$, and $C(2a + \sqrt{3}a, 5a)$.
To prove that the triangle is equilateral, we must show that the lengths of all three sides are equal.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of side $AB$:
$AB = \sqrt{(2a - 2a)^2 + (6a - 4a)^2} = \sqrt{0^2 + (2a)^2} = \sqrt{4a^2} = 2a$.
$2$. Length of side $BC$:
$BC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 6a)^2} = \sqrt{(\sqrt{3}a)^2 + (-a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
$3$. Length of side $AC$:
$AC = \sqrt{(2a + \sqrt{3}a - 2a)^2 + (5a - 4a)^2} = \sqrt{(\sqrt{3}a)^2 + (a)^2} = \sqrt{3a^2 + a^2} = \sqrt{4a^2} = 2a$.
Since $AB = BC = AC = 2a$, the triangle $ABC$ is an equilateral triangle.