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?
 A: Due to symmetry about the diagonal $AC$, $\angle NCB =( \angle DCB - \angle MCN) /2=15^{\circ}$
Hence $a=\sec 15^{\circ}$ and area of equilateral triangle is
$$\frac{\sqrt{3}}{4}\cdot \sec^2 15^{\circ}=2\sqrt{3}-3$$
A: 
Since the triangle is equilateral, the lengths of the sides must be equal; that is,
$$
2x^2=x^2-2x+2\tag1
$$
The solutions to $(1)$ are $x=-1\pm\sqrt3$. The only one that would apply to the case pictured above would be $-1+\sqrt3$. This would make the area of the triangle
$$
x-\tfrac12x^2=2\sqrt3-3\tag2
$$
A: 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 $x = 2-\sqrt{3}$.
A: 
Given square ABCD with AB=BC=CD=1, MNC an equilateral triangle MC=BC=MN=a.
For S$_{\triangle MNC}$, if we know MN=a, then we can directly apply S$_{\triangle MNC} =\frac{3}{4}a^2$.
Set NB = x, NC=MC=MN=a= $\sqrt{x^2+1}$, MD=$\sqrt{MC^2-1}$=x=NB, AM=AN=1-x.
Apply Pythagorean theroem to $\triangle ABN$
\begin{align} 2(1-x)^2=x^2+1
, x^2-4x+1=0, x=2\pm \sqrt{3} \end{align}
Since x <1, keep only  x=$2-\sqrt{3}$ $\to$ MN$^2$=a$^2$= 1+x$^2$=8-4$\sqrt 3$ \begin{align}
S_{\triangle MNC}=\frac{\sqrt{3}}{4}a^2=2\sqrt{3}-3 \end{align}
A: By symmetry
$$2 A + 60= 90, A= 15 ^{\circ}$$
From the narrow right triangle the hypotenuse
$$ a = \sec 15 ^{\circ}, Area=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4 \cos^215 ^{\circ}} =\frac{\sqrt{3}}{2 (1+\cos 30 ^{\circ})}=\sqrt3(2- \sqrt3). $$
