$P$ 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.
On
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
$$|\triangle ABC| = 2|\triangle PQR|$$
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 $2$.