Find the area of an equilateral triangle inscribed in a unit square If inside the square $ABCD$ there is an equilateral triangle $CMN$ inscribed, as shown in the figure. If the area of the square is 1, find the area of the triangle.

I attempted to say, I state that $x=MD$
then $MC^2=1+x^2$. $AM=1-x$ and I thought that something might come from Pythagoras, but it didn't. $\triangle AMN$ looks isosceles, but I can't prove that it is, although if it is proved to be isosceles, then the question can easily be solved by use of the pythagorean theorem. We also have that $\angle CNB+\angle CMD=150^o$. I got stuck here, could you please explain to me how to solve the question?

How about continuing to compute the lengths of the others segments?
You have seen that if $MD = x$, then $AM = 1-x$ and $MC = \sqrt{1+x^2}$. Now, since $MNC$ is equilateral, it follows that $MN = NC = \sqrt{1+x^2}$.
Applying Pythagoras theorem in the triangle $AMN$, we obtain: $AN = \sqrt{2x}$. It follows that $NB = 1- \sqrt{2x}$.
Now, we know all the lengths of the sides of the triangle $NBC$. Using again the Pythagoras theorem we have:
$$NC^2 = NB^2 + BC^2 \implies 1+x^2 = 1+2x-2\sqrt{2x}+1 \\ \implies 2\sqrt{2x} =-x^2+2x+1 \implies 8x = x^4+4x^2+1-4x^3-2x^2+4x \\ \implies x^4 -4x^3+2x^2-4x+1 = 0 \implies (x^2+1)(x^2-4x+1) = 0$$
The roots of this equation are $i, -i, 2-\sqrt{3}$ and $2+\sqrt{3}$. The only convenient value is $x = 2 - \sqrt{3}$.
Later edit: A simpler way of finding $x$ is to observe that the $\triangle MDC \equiv \triangle DBC$. Hence $MD = BN$, so $x = 1 - \sqrt{2x}$, thus $x^2 - 4x+1 = 0$ and from here you get $x = 2-\sqrt{3}$.
