In the parallelogram $ABCD$, the point $H$ is the closer to $A$ of the two trisection points of side $AB$, the point $F$ is the midpoint of side $BC$, and the point $G$ is the closest to $A$ of the two quarter points of side $DA$. In what ratio does the segment $HC$ divide the segment $FG$?
I. Solution. Let's try to project the ratios mentioned in the problem onto the line $GF$ using the theorem of parallel secants! Draw a line through $H$ parallel to the side $AD$ of the parallelogram, and let it intersect $GF$ at $E$ (Figure 1).

# 1986-11-382-1.eps 

Figure 1

Then

$$
\frac{GE}{EF}=\frac{AH}{HB}=\frac{1}{2}
$$

On the other hand, if $M$ denotes the intersection of $GF$ and $HC$, then

$$
\frac{EM}{MF}=\frac{EH}{FC}=\frac{EH}{FB}=\frac{2 \cdot GA + FB}{3 \cdot FB}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}
$$

We are essentially done, as we now know the ratio in which point $E$ divides segment $GF$ (1:2), and the ratio in which point $M$ divides segment $EF$ (2:3).

Now, let's choose $EM$ to be 4 units. Then, by (2), $MF$ is 6 units, so $EF$ is 10 units, and by (1), $GE$ is 5 units. From this, the desired ratio is:

$$
\frac{GM}{MF}=\frac{5+4}{6}=\frac{3}{2}
$$

Thus, segment $HC$ divides segment $FG$ in the ratio 2:3.

II. Solution. Let $AB = DC = 3a$ and $AD = BC = 4b$. Let the intersection of lines $GF$ and $HC$ be $M$, and the intersection of lines $HC$ and $AD$ be $P$ (Figure 2).

$$
\text{1986-11-383-1.eps}
$$

Figure 2

The triangles $PHA$ and $CHB$ are clearly similar, with the similarity ratio $AH:HB = 1:2$. Therefore, $AP = \frac{1}{2}BC = 2b$.

The triangles $CMF$ and $PMG$ are also similar, with the similarity ratio $CF:PG$, which is $CF = 2b$ and $PG = PA + AG = 2b + b = 3b$, so the ratio is $2:3$.

Thus, the ratio of $FM$ to $MG$ is also $2:3$, meaning that segment $HC$ divides segment $FG$ in the ratio 2:3.

Remark. Many made the following mistake: "The angles of quadrilaterals $GMCD$ and $FMHB$ are the same, so the two quadrilaterals are similar." This reasoning is incorrect! Consider, for example, a long, narrow rectangle and a square! Their angles are the same, but the two quadrilaterals are not similar.

For two quadrilaterals to be similar, in addition to the equality of angles, another condition is necessary, such as the ratio of two adjacent sides being the same. In our case, this is also true, because:

$$
GD:DC = 3b:3a = b:a = 2b:2a = FB:BH
$$

Thus, the two quadrilaterals are indeed similar. Using this, the problem can also be solved.

Two quadrilaterals are also similar if, when divided by one of their diagonals, the resulting triangles are pairwise similar.