points $R$ and $T$ lie on the side $CD$ of the parallelogram $ABCD$ such that $DR= RT= TC$ what is the area, in $cm^2$ , of the shaded region? points $R$ and $T$ lie on the side $CD$ of the parallelogram $ABCD$
such that
$DR= RT= TC$
. Lines $AR$ and $AT$ intersect the extension of $BC$ at
points $M$ and $L$ respectively, and the lines $BT$ and $BR$ intersect the extension of
$AD$ at points $S$ and $P$ respectively. If the area of the parallelogram $ABCD$ is $48$
cm$^2$
, then what is the area, in cm$^2$
, of the shaded region?

Im pretty sure ML=SP but im not sure how to prove it
 also I think $SMCD$ is a paralleogram but im not sure how to prove that either, hints or suggestions would be appreciated aswell as solutions
from the 2019 SAIMC
 A: Let $AD$ have the length of $2$ (units). Then $PD=1$ and $SP=3$. Same proportions on the segment $MB$. This is because we have:
$PD:PA=DR:AB=1:3$ and $SD:SA=DT:AB=2:3$.
This implies:
$$
\frac
{\operatorname{Area}(SPB)}
{\operatorname{Area}(DAB)}
=
\frac{SP}{DA}
=
\frac 32\ .
$$
The same also on the other side.
Now $\operatorname{Area}(DAB)$ is half of the area of the parallelogram.

Later edit: Thanks to the comment of 
ganeshie8
the above is not enough. The "common shaded area" is taken twice. So let us compute it. Let $X$ be the intersection of the diagonals of the parallelogram $SABM$. Let $Y$ be the intersection of the diagonals of the parallelogram $PABL$. Then $X$ is the mid point of the diagonals $SB$ and $MA$, as $P$, $L$ are also the mid points of the sides $SA$ and $MB$. So $XR:XA=PD:PA$ or $XR:XA=RT:AB$ and we get $1:3$ as a common value for $XR:XA=XT:XB$. We may need also the position of $Y$ on $RB$ and/or $AT$, for this we use the theorem of Menelaos for the triangle $RXB$ w.r.t. the line AYT:
$$
\frac{AR}{AX}\cdot
\frac{TX}{TB}\cdot
\frac{YB}{YR}
=1\ .
$$
This is:
$$
\frac23\cdot
\frac12\cdot
\frac{YB}{YR}
=1\ .
$$
So $YB:YR=3$.
Let $S$ be the area of the parallelogram $ABCD$.
We can finally write:
$$
\begin{aligned}
\operatorname{Area}(XAB)
&=
\frac 14
\operatorname{Area}(SABM)
=
\frac 14\cdot 3S
=
\frac 34S\ ,
\\
\operatorname{Area}(ARB)
&=
\operatorname{Area}(ARB)
=\frac 12S\ ,
\\
\operatorname{Area}(AXT)
&=
\operatorname{Area}(BXR)
\\
&=
\operatorname{Area}(AXB)
-
\operatorname{Area}(ATB)
=\left(\frac 34-\frac 12\right)S 
=\frac 14S\ ,
\\
\frac
{\operatorname{Area}(AYB)}
{\operatorname{Area}(ARB)}
&=
\frac{YB}{RB}
=\frac 34\ ,
\\
\operatorname{Area}(AYB)
&=
\frac 34
\operatorname{Area}(ARB)
=
\frac 34\cdot\frac 12S
=\frac 38S\ ,
\\
\operatorname{Area}(XRYT)
&=
\operatorname{Area}(AYB) +
\operatorname{Area}(AXT) +
\operatorname{Area}(BXR) -
\operatorname{Area}(AXB) 
\\
&=\left(\frac 14+\frac 14+\frac 38-\frac 34\right)S
\\
&=\frac 18S\ .
\end{aligned}
$$

Putting all together, the gray are is:
$$
\begin{aligned}
\operatorname{Area}(SPB) +
\operatorname{Area}(MLA) -
\operatorname{Area}(XRYT)
&=
\left(\frac 34+\frac 34-\frac 18\right)S
=
\frac {11}8S\ .
\end{aligned}
$$
A: Let $AL\cap BP=\{X\}$.
Thus, $$\frac{LC}{AD}=\frac{CT}{DT}=\frac{1}{2}=\frac{DR}{RC}=\frac{DP}{BC}=\frac{DP}{AD},$$ which gives
$$LC=DP,$$ $$AP=BL,$$ which gives $APLB$ is a parallelogram. 
By the similar way we obtain $$MC=2BC=2AD=DS,$$ which gives that $ASMB$ is a parallelogram.
Now, let $h$ be an altitude of $ABCD$ from $B$ to $AD$.
Thus, $$S_{shaded}=\frac{3}{4}S_{ASMB}-S_{\Delta APR}-S_{\Delta BLT}-S_{\Delta AXB}=$$
$$=\frac{3}{4}\cdot3AD\cdot h-2\cdot\frac{\frac{3}{2}AD\cdot\frac{1}{3}h}{2}-\frac{1}{4}S_{APLB}=$$
$$=\left(\frac{9}{4}-\frac{1}{2}\right)S_{ABCD}-\frac{1}{4}\cdot\frac{3}{2}AD\cdot h=\left(\frac{9}{4}-\frac{1}{2}-\frac{3}{8}\right)\cdot48=66.$$
