2. Let $M$ be the midpoint of side $B C$ of parallelogram $A B C D$ and let $A M$ intersect diagonal $B D$ at point $O$. Determine the area of quadrilateral $O M C D$, if the area of parallelogram $A B C D$ is $P$.
Solution. It is obvious that

$$
P_{A B M}=\frac{1}{4} P, P_{B C D}=\frac{1}{2} P
$$

and $\triangle B M O \sim \triangle A D O$ with a similarity coefficient of 2. From this, it follows that

$$
\overline{A O}: \overline{O M}=2: 1
$$

and

![](https://cdn.mathpix.com/cropped/2024_06_05_e95f2dbd17df8adc2c7cg-03.jpg?height=293&width=546&top_left_y=1887&top_left_x=919)

$$
\overline{A M}: \overline{O M}=3: 1
$$

Thus, $P_{A B M}=3 P_{B M O}$ (they have the same heights). Since $P_{A B M}=\frac{1}{4} P$, it follows that $P_{B M O}=\frac{1}{12} P$. Finally, we get

$$
P_{O M C D}=P_{B C D}-P_{B M O}=\frac{1}{2} P-\frac{1}{12} P=\frac{5}{12} P
$$