6. Calculate the ratio between the area of triangle $E C F$ and the area of square $A B C D$, if point $E$ is the midpoint of side $\overline{A B}$ and segment $\overline{E F}$ is perpendicular to diagonal $\overline{A C}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_24875f2c684caa0bffa1g-2.jpg?height=580&width=574&top_left_y=584&top_left_x=772)
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_4e3bea37fa3879b468d2g-6.jpg?height=394&width=396&top_left_y=294&top_left_x=296)

Let $a$ be the length of the side of square $ABCD$.

Since point $E$ is the midpoint of side $\overline{AB}$, it follows that $|AE| = \frac{a}{2}$.

In the right triangle $AEF$, $|\angle FAE| = 45^{\circ}$, so triangle $AEF$ is also isosceles. 1 POINT

Let $|AF| = |FE| = x$.

Applying the Pythagorean theorem in triangle $AEF$, we get $x = \frac{a \sqrt{2}}{4}$.

The length of the diagonal of square $ABCD$ is $a \sqrt{2}$, i.e., $|AC| = a \sqrt{2}$.

1 POINT

Furthermore, $|CF| = a \sqrt{2} - \frac{a \sqrt{2}}{4} = \frac{3}{4} a \sqrt{2}$.

The area of the right triangle $ECF$ is $P_{EFC} = \frac{\frac{3}{4} a \sqrt{2} \cdot \frac{a \sqrt{2}}{4}}{2} = \frac{\frac{3}{8} a^2}{2} = \frac{3}{16} a^2$,

The area of square $ABCD$ is $a^2$.

The ratio of the areas of triangle $ECF$ and square $ABCD$ is $3: 16$.

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