10.106. On the sides of an isosceles right triangle with hypotenuse outside this triangle, squares are constructed. The centers of these squares are connected to each other. Find the area of the resulting triangle.
Solution.

$B E=K C=\frac{c}{2} ; E F=2 B E=c$ (Fig. 10.102). Quadrilateral $A B C D$ is a square. Therefore, $A C=B D=c$. The area of the desired triangle is

$$
S=\frac{1}{2} B D \cdot E F=\frac{1}{2} c^{2}=\frac{c^{2}}{2}
$$

Answer: $\frac{c^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-607.jpg?height=404&width=420&top_left_y=94&top_left_x=172)

Fig. 10.103

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-607.jpg?height=420&width=468&top_left_y=94&top_left_x=734)

Fig. 10.104