Find the ratio of areas in the $ABP$ and $PMCQ$ regions. For reference: In the figure, $(BM):(AQ)=(MC):(QC)$. calculate the ratio
of areas in the $ABP$ and $PMCQ$ regions.

My progress
$\frac{S_{ABP}}{S_{PMCQ}}=?\\
\frac{x}{y}=\frac{m}{n}\\
\\
\frac{S_{ABM}}{S_{AMC}}=\frac{x}{y}\\
\frac{S_{ABQ}}{S_{BQC}}=\frac{m}{n}\\
\frac{S_{ABM}}{S_{AMC}}=\frac{S_{ABQ}}{S_{BQC}}$
I couldn't develop from here.
 A: A couple of tricks which are useful for similar problems:

*

*Affine maps preserve the ratios of areas and the ratios of lengths of segments on the same line, so you may assume without loss of generality that $\widehat{C}=90^\circ,AC=CB$ and $[ABC]=1$, for instance;

*If we draw the diagonals of a trapezoid, there are important relations about the areas of the four resulting triangles.


The hyphotesis gives $MQ\parallel AB$, hence $CP$ meets $AB$ at its midpoint by (the converse of) Ceva's theorem. Let $\lambda=CQ/CA=CM/CB$. By Van Obel's theorem it follows that $BP/PQ=\frac{1-\lambda}{\lambda}+1=\frac{1}{\lambda}$. This leads to
$$ [AMB]=(1-\lambda)[ABC]=(1-\lambda),\qquad [APB]=\frac{1}{\lambda+1}[AMB]=\frac{1-\lambda}{1+\lambda}$$
$$[BMP]=[AQP]=\frac{\lambda}{1+\lambda}[AMB]=\frac{\lambda(1-\lambda)}{(1+\lambda)}$$
$$ [CQPM]=[ABC]-[AMB]-[AQP]= \lambda-\frac{\lambda(1-\lambda)}{1+\lambda}=\frac{2\lambda^2}{1+\lambda}$$
so
$$\frac{[APB]}{[CQPM]} = \frac{1-\lambda}{2\lambda^2}. $$
A: 
Given, $\frac{BM}{AQ}=\frac{MC}{QC}$
so  $$\frac{BM}{MC}=\frac{AQ}{QC}=\frac{λ}{1}$$ (say)
 By Menelaus’ Theorem on triangle BQC,
 $$\frac{AC}{AQ}\frac{QP}{PB}\frac{BM}{MC}=1$$
which implies that $\frac{QP}{PB}=\frac{1}{λ+1}$.Similarly by Menelaus’ Theorem on triangle AMC, $\frac{MP}{AP}=\frac{1}{λ+1}$.
 By the property that ratio of areas of triangles with same height is equal to the ratio of their bases, we have $$\frac{ar(ABP)}{ar(APQ)}=\frac{λ+1}{1}$$,
 $$\frac{ar(ABP)}{ar(MBP)}=\frac{λ+1}{1}$$
 So ar(APQ)=ar(MBP). Let ar(APB)=(λ+1)k, then ar(APQ)=ar(MBP)=k. Now,
 $$\frac{ar(ABQ)}{ar(BQC)}=\frac{λ}{1}$$
 $$\frac{(λ+2)k}{k+ar(PMCQ)}=\frac{λ}{1}$$
 $(λ+2)k= λk+λ.ar(PMCQ)$ so that ar(PMCQ)=$\frac{2k}{λ}$. This implies that
$$\frac{ar(ABP)}{ar(PMCQ)}=\frac{λ(λ+1)}{2}=\frac{AQ•AC}{2QC^2}=\frac{BM•BC}{2MC^2}$$ where $\frac{AQ}{QC}=\frac{λ}{1}$ .
