5. In square $A B C D$, let $K$ be the midpoint of side $A B$ and $L$ be the midpoint of side $A D$. Segments $K D$ and $L C$ intersect at point $M$ and divide the square into two triangles and two quadrilaterals. Calculate their areas if segment $L M$ has a length of $1 \mathrm{~cm}$.
SOLUTION. It holds that $|A K|=|D L|$ and $|A D|=|D C|=2|A K|$ (Fig. 2), so the right triangles $A K D$ and $D L C$ are congruent by the SAS theorem. In addition, triangles $M L D$ and $A K D$ are similar by the AA theorem, because $|\angle L D M|=|\angle K D A|$ and $|\angle D L M|=$ $=|\angle D L C|=|\angle A K D|$. Similarly, the similarity of triangles $M D C$ and $A K D$ can be verified. From the similarity of triangles $A K D, M L D$, and $M D C$, it follows that $|M D|=2|M L|=2 \mathrm{~cm}$ and $|M C|=2|M D|=4 \mathrm{~cm}$. The areas of figures $M L D, M D C$, and $A K M L$ are

$$
S_{M L D}=\frac{1 \cdot 2}{2}=1 \mathrm{~cm}^{2}, \quad S_{M D C}=\frac{2 \cdot 4}{2}=4 \mathrm{~cm}^{2}
$$

and

$$
S_{A K M L}=S_{A K D}-S_{M L D}=S_{D L C}-S_{M L D}=S_{M D C}=4 \mathrm{~cm}^{2}
$$

Finally, using the Pythagorean theorem, we get $S_{A B C D}=|D C|^{2}=|D M|^{2}+|C M|^{2}=$ $=20 \mathrm{~cm}^{2}$, so

$$
S_{K B C M}=S_{A B C D}-\left(S_{M L D}+S_{M D C}+S_{A K M L}\right)=11 \mathrm{~cm}^{2} .
$$

Conclusion. The areas of triangles $M L D, M D C$ and quadrilaterals $A K M L, K B C M$ are, respectively, $1 \mathrm{~cm}^{2}, 4 \mathrm{~cm}^{2}, 4 \mathrm{~cm}^{2}$, and $11 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8ac08bf99a5225785e64g-3.jpg?height=480&width=551&top_left_y=1890&top_left_x=410)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_04_17_8ac08bf99a5225785e64g-3.jpg?height=482&width=534&top_left_y=1889&top_left_x=1112)

Fig. 3

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