## Solution

Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.

Then by Pythagorean Theorem we can prove that $M A^{2}+M C^{2}=M B^{2}+M D^{2}$.

From the given condition we obtain $M A=1$.

With center $A$ and angle $60^{\circ}$, we rotate $\triangle A M E$, so we construct the triangle $A N B$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-5.jpg?height=394&width=491&top_left_y=360&top_left_x=774)

Since $A M=A N$ and $\widehat{M A N}=60^{\circ}$, it follows that $\triangle A M N$ is equilateral and $M N=1$. Hence $\triangle B M N$ is right-angled because $B M^{2}+M N^{2}=B N^{2}$.

So $m(\widehat{B M A})=m(\widehat{B M N})+m(\widehat{A M N})=150^{\circ}$.

Applying Pythagorean Generalized Theorem in $\triangle A M B$, we get:

$A B^{2}=A M^{2}+B M^{2}-2 A M \cdot B M \cdot \cos 150^{\circ}=1+2+2 \sqrt{2} \cdot \sqrt{3}: 2=3+\sqrt{6}$.

We conclude that the area of the square $A B C D$ is $3+\sqrt{6}$.


G5 Inside the square $A B C D$, the equilateral triangle $\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\triangle A B E$ such that $M B=\sqrt{2}, M C=\sqrt{6}, M D=\sqrt{5}$ and $M E=\sqrt{3}$. Find the area of the square $A B C D$.

