Prove that the points A(2, 4), B(2, 6) and

The given points are $A(2,4), B(2,6)$ and $C(2+\sqrt{3}, 5)$. Now

$A B=\sqrt{(2-2)^{2}+(4-6)^{2}}=\sqrt{(0)^{2}+(-2)^{2}}$

$=\sqrt{0+4}=2$

$B C=\sqrt{(2-2-\sqrt{3})^{2}+(6-5)^{2}}=\sqrt{(-\sqrt{3})^{2}+(1)^{2}}$

$=\sqrt{3+1}=2$

$A C=\sqrt{(2-2-\sqrt{3})^{2}+(4-5)^{2}}=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}$

$=\sqrt{3+1}=2$

Hence, the points $A(2,4), B(2,6)$ and $C(2+\sqrt{3}, 5)$ are the vertices of an equilateral triangle.