Find the area of ​the $AMRQ$ region For reference: In figure : $ABCD$, it is a parallelogram
$MN \parallel AD, PQ \parallel AB$; if: the area of $​​RPCN$ is $20m^2$
calculate the area of ​​the $AMRQ$ region (Answer:$20m^2$)

My progress:
$BM=a\\
AM=b\\
BP=m\\
PC=n\\
S_{MRA}=A,S_{ARQ}=B,S_{PRC}=M, S_{CRN}=N\\
A+B=x\\
M+N=20\\
S_{BMR} = \frac{Aa}{b}\\
S_{DRQ}=\frac{Bn}{m}\\
S_{NDR}=\frac{Nb}{a}\\
S_{BPR}=\frac{Mm}{n}\\
S_{ABD}=S_{BCD}\\
x+\frac{Aa}{b}+\frac{Bn}{m}=20+\frac{Mm}{n}+\frac{Nb}{a}$
...???

 A: Hint: $QFND$ is a parallelogram so altitude from $Q$ and $N$ to $DF$ is equal and given the common base $DR$, $S_{\triangle RQD} = S_{\triangle RND}$.
Similarly, $BMFP$ is a parallelogram and $S_{\triangle BRM} = S_{\triangle BRP}$
A: BPFM and FNQD are paralelograms
$ML = LP, QT=TN\\
\therefore S_{RBM}=S_{RBP}\\
S_{RQF}=S_{RNF}\\
20+A+C+2B =S+A+C+2B\\
\therefore S = 20$

A: Transform the parallelogram into a unit square, then the ratio of areas is preserved.
The images of the various points will be as follows
$A = (0, 0)$
$B = (0, 1)$
$C = (1,1)$
$D = (1, 0)$
$R = (s, 1- s)$
$Q = (t, 0)$
$P = (t, 1)$
$N = (1, r)$
$M = (0, r)$
$F = (t, r)$
Since point $F$ lies on the diagonal, then $r = 1- t$
Now apply the so-called shoelace formula, to both quadrilaterals
$[PRCN] = (1, r) \times (1,1) + (1, 1) \times (t, 1) + (t, 1) \times (s, 1- s) + (s, 1- s) \times (1, r) $
where $(a, b ) \times (c, d) = a d - b c $
This reduces to
$[PRNC] = 1- r - ts + s r$
Similarly calculation for [AMRQ] results in
$[AMRQ] = (t, 0) \times (s, 1- s) + (s, 1 - s) \times (0, r) = t - t s + s r $
But as was stated above $r = 1 - t$ , so $t = 1 - r$ and therefore the two areas are equal.
