Prove, square of quadrilateral is the sum of squares of 4 triangles Let $A_1$, $B_1$, $C_1$, and $D_1$ - midpoints of the sides $AB$, $BC$, $CD$ and $DA$ convex quadrilateral $AВСD$. Directs $AC_1$, $ВD_1$, $CA_1$ and $DВ_1$ - divide it by $5$ quadrilaterals and $4$ triangles. Prove that the area of ​​the central quadrilateral is the sum of squares of $4$ triangles.  
I think, it should be proved in such a way: the square of quadrilateral is finding out by such formula: $S=(p-a)(p-b)(p-c)-abcd\cos^2\left(\frac{A+B}{2}\right)$. If we will prove that the sum of big triangles are the same as the square of $ABCD$, it will be automatically proved area of the central quadrilateral is the sum of areas of small triangles.


Since $A_1A=A_1B$, $~C_1C=C_1D$, then 
$$
S_{A_1BC} = S_{AA_1C}, \qquad S_{C_1DA} = S_{CC_1A};
$$
since $D_1D=D_1A$, $~B_1B=B_1C$, then 
$$
S_{D_1AB} = S_{DD_1B}, \qquad S_{B_1CD} = S_{BB_1D};
$$
so,
$$
S_{A_1BC}+S_{B_1CD}+S_{C_1DA}+S_{D_1AB} = S_{AA_1C}+S_{BB_1D}+S_{CC_1A}+S_{DD_1B},
$$
other words,
$$
\color{green}{S_{\large{green}}} + 2\cdot \color{blue}{S_{\large{blue}}} =
\color{green}{S_{\large{green}}} + 2\cdot \color{red}{S_{\large{red}}};
$$
$$
\color{blue}{S_{\large{blue}}} =
\color{red}{S_{\large{red}}}.
$$
