Find the ratio of the area of two triangles $P$ and $Q$ are the midpoints of $AB$ and $BC$ respectively. $S$ and $T$ and the midpoints of $PR$ and $QR$. Find the ratio of area of $\triangle ABC$ and $\triangle PQR$

Don't forget to include your approach towards solving the question. Things to look out for.
My work. I can figure out ratio of area of $\triangle RST / \triangle RPQ = 1/4$. 
Same for the ratios of the area of $\triangle BQP/\triangle BCA =1/4$. Can't figure out what to do next. 
 A: Note that $|ST|=\frac{1}{2}|PQ|=\frac{1}{4}|AC|$ (see  G Cab's comment), and 
$$\frac{|\triangle ASP|}{|AS|}=\frac{|\triangle STR|}{|ST|}=\frac{|\triangle TCQ|}{|TC|}$$
(try to figure out why these equalities hold).
Therefore, since $|\triangle STR|=\frac{1}{4}|\triangle PQR|$, it follows
\begin{align}
|\triangle ASP|+|\triangle TCQ|
&=\frac{|AS|+|TC|}{|ST|}\cdot |\triangle STR|\\
&=\frac{|AC|-|ST|}{|ST|}\cdot |\triangle STR|\\
&=3|\triangle STR|=\frac{3}{4}|\triangle PQR|.
\end{align}
Finally $|\square PQCA|=\frac{3}{4}|\triangle ABC|$
and $|\square PQTS|=\frac{3}{4}|\triangle PQR|$, yield
$$\frac{3}{4}|\triangle ABC|=|\square PQCA|=(|\triangle ASP|+|\triangle TCQ|)
+|\square PQTS|=\frac{3}{2}|\triangle PQR|$$
that is the ratio $|\triangle ABC|/|\triangle PQR|$ is equal to $2$.
A: By the Midsegment Theorem, we know that $\overline{PQ}\parallel\overline{AC}$. Let's draw a couple more parallels, through $B$ and $R$.

Recall that families of parallel lines cut transversals in equal ratios. Since $\overleftrightarrow{PQ}$, $\overleftrightarrow{AC}$, $\overleftrightarrow{A^\prime C^\prime}$ cut $\overline{PR}$ (and $\overline{QR}$) equally, they cut $\overline{PA}$ and $\overline{QC}$ equally. We conclude that all four lines are equally-spaced, creating a trisected perpendicular $\overline{A^\prime B^\prime}$.
Thus, $\triangle ABC$ and $\triangle PQR$ have equal heights relative to bases $\overline{AC}$ and $\overline{PQ}$. Since, again by the Midsegment Theorem, $|\overline{AC}| = 2|\overline{PQ}|$, necessarily

$$|\triangle ABC| = 2|\triangle PQR|$$

