Step 1 :

Let $A(2,3)$ be one vertex and $x+y=2$ be the opposite side of the given equilateral triangle.

Hence the other two sides pass through the point A(2,3) and make an angle $60^{\circ}$

Slope of the line $x+y=2$ is 1.

$m = -1$

$ \therefore $ Equation of the other two sides are

$ y-3 = \bigg( \large\frac{-1- \tan 60^{\circ}}{1-\tan 60^{\circ}} \bigg) $$(x-2)$

and $y-3 = \bigg( \large\frac{-1-\tan 60^{\circ}}{1+ \tan 60^{\circ}} \bigg) $$(x-2)$

But $\tan 60^{\circ} = \sqrt 3 $

$ \therefore y-3 = \large\frac{-1-\sqrt 3}{1-\sqrt 3 }$$(x-2)$

(i.e) $y-3=\large\frac{-(1+ \sqrt 3 )}{1-\sqrt 3}$$(x-2)$

(i.e) $ (y-3) = (2+\sqrt 3)(x-2)$

is the equation of one side.

$y-3 = \large\frac{\sqrt 3 - 1 }{\sqrt 3 + 1 }$$(x-2)$

$ \Rightarrow (y-3)(2+ \sqrt 3 )=(x-2)$ is the equation of the other side.

(i.e) $(y-3)=(2-\sqrt 3 )(x-2)$

Hence the statement is true.