## Task 5 - 300735

Let $A B C D$ be a rectangle with side lengths $A B=a$ and $B C=b$, and let $a>b$. On $A B$, points $C_{1}$ and $D_{1}$, and on $C D$, points $A_{1}$ and $B_{1}$ are marked such that the segments $A A_{1}, B B_{1}, C C_{1}, D D_{1}$ are the angle bisectors of an interior angle of $A B C D$.

The intersection points $E, F, G, H$ of these angle bisectors are labeled as shown in the figure.

Determine the area $I$ of the quadrilateral $E F G H$, given that $a=8 \mathrm{~cm}$ and $b=5 \mathrm{~cm}$!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0527.jpg?height=346&width=443&top_left_y=358&top_left_x=1292)
From $A A_{1}, B B_{1}, C C_{1}, D D_{1}$, which bisect the right angles at $A, B, C$ and $D$ respectively, it follows:

The triangles $A A_{1} D, B B_{1} C, C C_{1} B, A D D_{1}$ each have two interior angles of $90^{\circ}, 45^{\circ}$. Therefore, the third interior angle in each triangle is also $45^{\circ}$; the triangles are isosceles right triangles with the leg length $b$.

Thus, $A D_{1} A_{1} D$ and $C_{1} B C B_{1}$ are squares with side length $b$. Since their diagonals are equal in length and bisect each other, it follows that $A H=A 1 H=B F=B 1 F=C F=C 1 F=D H=D 1 H$.

Furthermore, $A D H, B C F$ as well as $C_{1} D_{1} E$ and $A_{1} B_{1} G$ are triangles with two interior angles of $45^{\circ}$ each, thus they are also isosceles right triangles.

Here, $C_{1} D_{1}=A B-B D_{1}=a-2(a-b)=2 b-a$, and $A_{1} B_{1}$ has the same length; consequently, $C_{1} E=D_{1} E=A_{1} G=B_{1} G$. Therefore, $E F G H$ is a quadrilateral with four interior angles of $90^{\circ}$ and with $H E=H G=F E=F G$. That is, a square.

The length of its diagonals is $d=H F=A C_{1}=a-b=3 \text{ cm}$, since $A H$ and $C_{1} F$ are equal in length and parallel, thus $A C_{1} F H$ is a parallelogram. Since in the square $E F G H$ the diagonals are equal in length and perpendicular to each other and bisect each other, the perpendicular height to $H F$ in each of the two triangles $H F E, H F G$ is $\frac{d}{2}$. The sum of the areas of these two triangles, i.e., the area of $E F G H$, is therefore

$$
I=\frac{1}{2} d^{2}=4.5 \text{ cm}^{2}
$$