Prove that $S_{ABC}=2S_{MNP}$ We draw a line out of triangle $ABC$ and then we draw prepediculars $AA',BB',CC'$ to that line and name the mid point of those $M,N,P$.Prove that $S_{ABC}=2S_{MNP}$
We have midpoints and some trapezoid it is obvious that the area of that trapezoids becomes half but I don't know the rest.
Edit:$S$ is the area of the triangle.
 A: 
The area of the $\triangle ABC$ is expressed 
in terms of coordinates $A(A_x,A_y),B(B_x,B_y),C(C_x,C_y)$
as
\begin{align}
S_{ABC}&=\tfrac12|(B_x-A_x)(C_y-A_y)-(C_x-A_x)(B_y-A_y)|
,\\
\text{similarly, }
S_{MNP}&=\tfrac12|(N_x-M_x)(P_y-M_y)-(P_x-M_x)(N_y-M_y)|
.
\end{align}
Let's consider the line $\mathcal{L}$ as the $x$-axis,
then 
\begin{align}
A_x&=M_x,\quad B_x=N_x, \quad C_x=P_x
,\\
A_y&=2\,M_y,\quad B_y=2\,N_y,\quad C_y=2\,P_y
.
\end{align}
Then
\begin{align}
S_{ABC}&=\tfrac12|(N_x-M_x)(2\,P_y-2\,M_y)-(P_x-M_x)(2\,N_y-2\,M_y)||
=2\,S_{MNP}
.
\end{align}
A: $$
S_{ABC}=S_{A'ACC'}+S_{B'BCC'}-S_{A'ABB'}=
2(S_{A'MPC'}+S_{B'NPC'}-S_{A'MNB'})=2S_{MNP}
$$
[Diagram from g.kov's answer]

A: Let us take line $\frak{L}$ as the support of a first coordinate axis. Let the ordinate line be any line orthogonal to $\frak{L}$. Then triangle $MNP$ is the image of triangle $ABC$ by the affine transform given by matrix:
$$T=\left(\begin{array}{cc}1 & 0\\ 0 & \tfrac12 \end{array}\right).$$
It is well know that, whatever the figure $A$, the ratio of areas $\frac{S_{T(A)}}{S_A}$ is given by the determinant of transform $T$, i.e., $\tfrac12$.
