5. Given the quadrilaterals $A B C D$ and $E F G H$, such that the points $B, C, D$ and $A$ are the midpoints of the segments $A E, B F, C G$ and $D H$, respectively. Determine the area of quadrilateral $E F G H$, if the area of quadrilateral $A B C D$ is $1 \mathrm{~cm}^{2}$.
Solution. Let $P_{\triangle A B D}=P_{1}$ and $P_{\triangle B C D}=P_{2}$. For triangles $A B H$ and $A B D$, we have $\overline{A D}=\overline{A H}$ and they share a common height from vertex $B$, so $P_{\triangle A B H}=P_{\triangle A B D}$. For triangles $H B E$ and $H A B$, we have $\overline{A B}=\overline{B E}$ and they share a common height from vertex $H$, so $P_{\triangle H B E}=P_{\triangle H A B}$. Therefore,

$$
P_{\triangle A H E}=2 P_{\triangle A B H}=2 P_{\triangle A B D}=2 P_{1}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6d6f67207af7682f8d1dg-8.jpg?height=713&width=933&top_left_y=999&top_left_x=471)

Similarly, it can be shown that $P_{\triangle C F G}=2 P_{2}$.

Let $P_{\triangle A B C}=P_{3}$ and $P_{\triangle A C D}=P_{4}$. In a completely analogous manner, as in the previous considerations, it can be shown that $P_{\triangle B E F}=2 P_{3}$ and $P_{\triangle D G H}=2 P_{4}$. Finally, the area of quadrilateral $E F G H$ is equal to the sum of the areas of triangles $A E H, B E F, C F G, D G H$ and quadrilateral $A B C D$, so if we use the fact that $P_{1}+P_{2}=P_{3}+P_{4}=1 \mathrm{~cm}^{2}$, we get

$$
P=2 P_{1}+2 P_{2}+2 P_{3}+2 P_{4}+1=2\left(P_{1}+P_{2}\right)+2\left(P_{3}+P_{4}\right)+1=2+2+1=5 \mathrm{~cm}^{2}
$$