Let the centroid of triangle $ABC$ be $S$, the midpoint of segment $AS$ be $H$, and the midpoint of side $AB$ be $Q$. The line through $H$ parallel to $BC$ intersects side $AB$ at $P$ and line $CQ$ at $R$. What is the ratio of the areas of triangles $PQR$ and $APH$?
Let $A S$ be the line and $B C$ the segment intersecting at point $F$. Since $S$ is the centroid of triangle $A B C$, $F$ is the midpoint of segment $B C$. Since $P H$ is parallel to $B F$, triangles $A P H$ and $A B F$ are similar. The similarity ratio is $1: 3$, because $H$ bisects segment $A S$, and $S$ trisects segment $A F$, thus

$$
A H=\frac{1}{2} A S=\frac{1}{2} \cdot \frac{2}{3} A F
$$

Therefore, $P H=\frac{1}{3} B F$. Triangles $R H S$ and $C F S$ are congruent because their corresponding sides are parallel and $H S = S F = \frac{1}{3} A F$. Therefore,

$$
R H=C F=\frac{1}{2} B C=3 \cdot P H
$$

Thus, $P$ is farther from vertex $R$ and trisects the median $R H$ of triangle $A R S$, so $P$ is the centroid of triangle $A R S$.

Accordingly, segments $A Q$ and $R H$ are medians of triangle $A R S$, which means the areas of triangles $P Q R$ and $A P H$ are equal.

Based on the work of Puskás Anna (Fazekas M. Főv. Gyak. Gimn., 9th grade)

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