Find the area of triangle $RAQ$ in the figure below

Question

For reference:

Through the midpoint $M$ of the side $AC$ of a triangle $ABC$, a line is drawn that intersects in $BC$ in $P$ and extended $BA$ in $Q$. By $B$, a line parallel to $PQ$ is drawn which intersects the extension of $CA$ at $R$. If the area of ​​the $RPC$ region is $12\ \mathrm{m^2}$ calculate the area of ​​the $RAQ$ region.(Answer: $12\ \mathrm{m^2}$)

My progress:

Follow the relations I found Let $D=RP \cap BQ \implies [DQR]=[BDP]$

$\dfrac{[ABR]}{[ABC]}=\dfrac{AR}{AC}$

$\dfrac{[AQR]}{[ABC]}=\dfrac{AR\cdot AQ}{AB\cdot AC}$

$\dfrac{[AQR]}{[AQM]}=\dfrac{AR}{AM}$

$\dfrac{12}{[CMP]}=\dfrac{CR}{CM}$

$\dfrac{12}{[AMP]}=\dfrac{CP}{MP}$

...???

Answer 1

Connect $B$ and $M$.

We have $$[RAQ] = [BAM] = [BCM] = [BPM] + [CPM] = [RPM] + [CPM] = [RPC].$$

Answer 2

In $\triangle ABC$, using secant $PQ$ and applying Menelaus's theorem,

$\displaystyle \frac{BQ}{AQ} \cdot \frac{AM}{MC} \cdot \frac{CP}{PB} = 1 \implies \frac{[RCP]}{[RPB]} = \frac{[RAQ]}{[RBQ]}$

But as $PQ \parallel BR$ and $\triangle RPB$ and $\triangle RBQ$ have common base $BR, [RPB] = [RBQ]$.

So $[RAQ] = [RCP] = 12$