In the figure below, $ABCD$ is a parallelogram, and $M$ is the midpoint of segment $AD$.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-17.jpg?height=562&width=1211&top_left_y=1775&top_left_x=525)

If the area of quadrilateral $MOCD$ is equal to $5 \mathrm{~cm}^{2}$, calculate the area of triangle $AOM$.

#
Solution

Segments $A D$ and $B C$ are parallel. Therefore, the following equalities between the angles hold: $O \hat{A} M=O \hat{C} B$ and $O \hat{B} C=O \hat{M} A$.

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-47.jpg?height=397&width=811&top_left_y=664&top_left_x=725)

Thus, triangles $A O M$ and $C O B$ are similar (that is, these triangles have the same internal angles). Since point $M$ is the midpoint of segment $A D$, we have that

$$
\overline{B C}=\overline{A D}=2 \times \overline{A M}
$$

Therefore, the ratio between triangles $A O M$ and $C O B$ is 1 to 2, that is, the length of each side of triangle $C O B$ is twice the length of its corresponding side in triangle $A O M$. This shows that $\overline{B O}=2 \times \overline{O M}$.

Note that segment $A H$ functions as a common height for triangles $B A O$ and $O A M$ as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-47.jpg?height=394&width=808&top_left_y=1645&top_left_x=724)

Thus, we conclude that

$$
\text { Area }(\triangle B A O)=\frac{\overline{B O} \times \overline{A H}}{2}=2 \times \frac{\overline{O M} \times \overline{A H}}{2}=2 \times \operatorname{Area}(\triangle O A M)
$$

Similarly, triangles $B A O$ and $O B C$ share the same height. Since $\overline{O C}=2 \times \overline{A O}$, then

$$
\text { Area }(\triangle O B C)=2 \times \text { Area }(\triangle A O B)
$$

Let's call $a$ the area measure of triangle $A O M$. Above, we have just shown that the area of triangle $A O B$ is equal to $2a$, and that of triangle $O B C$ is equal to $4a$, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_27f67f4520ebb8354fb2g-48.jpg?height=389&width=806&top_left_y=231&top_left_x=525)

Now, observe that triangles $A B C$ and $A D C$ are congruent (that is, they have the same side lengths). Therefore, these two triangles have the same area. Since the area of triangle $A B C$ is equal to $6a$, for triangle $A C D$ to have the same area, it is necessary that the area of quadrilateral $M O C D$ be equal to $5 \times a$. From the problem statement, we have that $5a = 5 \text{ cm}^2$. Therefore, $a = 1 \text{ cm}^2$. The answer is then $1 \text{ cm}^2$ of area for triangle $A O M$.