2. On the sides $AB$ and $BC$ of parallelogram $ABCD$, points $K$ and $M$ are chosen, respectively, such that $AK: KB = 2: 3$ and $BM: MC = 2: 1$. Find the ratio of the areas of $\triangle BKM$ and $\triangle DKM$, i.e., calculate the value $\frac{P(\triangle BKM)}{P(\triangle DKM)}$.
2. First solution. Let $S$ be the intersection of the diagonal $DB$ and the segment $KM$. Since $\triangle BKM$ and $\triangle DKM$ share the side $KM$, the ratio of their areas is actually the ratio of their heights dropped onto $KM$. The heights dropped from $B$ and $D$ to $KM$ are in the ratio of $SB$ to $DS$ (by Thales' theorem), so we need to find $\frac{SB}{DS}$.

Let $\vec{x} = \overrightarrow{DC}$ and $\vec{y} = \overrightarrow{DA}$. Then from $\overrightarrow{DB} = \vec{x} + \vec{y}$ and $\overrightarrow{DS} = k \overrightarrow{DB}$ for some $k, 0 < k < 1$ (and thus $\overrightarrow{SB} = \overrightarrow{DB} - \overrightarrow{DS} = (1 - k) \overrightarrow{DB}$), we have

$$
\overrightarrow{DS} = k \vec{x} + k \vec{y}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_841f52b4ffb037ad799bg-51.jpg?height=254&width=433&top_left_y=856&top_left_x=1074)

On the other hand, since $S$ is on the segment $KM$, for some real number $m, 0 < m < 1$, we have $\overrightarrow{DS} = m \overrightarrow{DK} + (1 - m) \overrightarrow{DM}$. Given $\overrightarrow{DK} = \overrightarrow{DA} + \overrightarrow{AK} = \vec{y} + \frac{2}{5} \vec{x}$ and $\overrightarrow{DM} = \overrightarrow{DC} + \overrightarrow{CM} = \vec{x} + \frac{1}{3} \vec{y}$, substituting these into the previous equation gives

$\overrightarrow{DS} = m \left( \vec{y} + \frac{2}{5} \vec{x} \right) + (1 - m) \left( \vec{x} + \frac{1}{3} \vec{y} \right) = \left( \frac{2m}{5} + 1 - m \right) \vec{x} + \left( m + \frac{1 - m}{3} \right) \vec{y} = \frac{5 - 3m}{5} \vec{x} + \frac{2m + 1}{3} \vec{y}$.

Thus, we have $k \vec{x} + k \vec{y} = \frac{5 - 3m}{5} \vec{x} + \frac{2m + 1}{3} \vec{y}$. Since the vectors $\vec{x}$ and $\vec{y}$ are non-collinear, it must be that $k = \frac{5 - 3m}{5}$ and $k = \frac{2m + 1}{3}$. Therefore, $\frac{5 - 3m}{5} = \frac{2m + 1}{3}$, which simplifies to $15 - 9m = 10m + 5$, and solving for $m$ gives $m = \frac{10}{19}$. Finally, $k = \frac{2 \cdot \frac{10}{19} + 1}{3} = \frac{13}{19}$ and

$$
\frac{P(\triangle BKM)}{P(\triangle DKM)} = \frac{SB}{DS} = \frac{1 - k}{k} = \frac{\frac{6}{19}}{\frac{13}{19}} = \frac{6}{13}
$$

Second solution. As in the previous solution, we seek $\frac{SB}{DS}$. Draw a line through $M$ parallel to $AB$, and let $P$ be the intersection of this line with $BD$. Considering the lines $BD$ and $BC$, by Thales' theorem we have $\frac{MP}{CD} = \frac{MB}{CB} = \frac{2}{3}$, i.e., $MP = \frac{2}{3} CD$, and $\frac{DP}{DB} = \frac{CM}{CB} = \frac{1}{3}$, i.e., $DP = \frac{1}{3} DB$. Considering the lines $BD$ and $KM$, by Thales' theorem we have $\frac{BS}{PS} = \frac{BK}{PM} = \frac{\frac{3}{2}}{\frac{2}{3} CD} = \frac{9}{10}$. Let $BS = 9x$ and $PS = 10x$. We have $BS + PS = BP = DB - DP = \frac{2}{3} DB$, i.e., $19x = \frac{2}{3} DB$, so $x = \frac{2}{57} DB$. Therefore, $BS = 9x = \frac{6}{19} DB$ and $DS = DB - BS = \frac{13}{19} DB$, and finally

$$
\frac{P(\triangle BKM)}{P(\triangle DKM)} = \frac{SB}{DS} = \frac{\frac{6}{19} DB}{\frac{13}{19} DB} = \frac{6}{13}
$$