In the symmetric trapezoid $ABCD$, the lengths of the parallel sides are $a$ and $c$. Let the midpoints of the legs be $E$ and $F$. The perpendicular projection of point $E$ onto the line of leg $BC$ is point $G$. What is the area of the trapezoid if point $C$ is the closer third point of segment $GF$ to $G$?
Solution. From the text of the problem, it follows that points $E$ and $F$ are the midpoints of the legs $AD$ and $BC$, respectively, and that the base $AB$ is longer than the base $CD$. Therefore, we can choose the notation such that $a > c$ holds.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e48b51ccbfb5066fcd0g-1.jpg?height=407&width=596&top_left_y=268&top_left_x=754)

Let the length of the legs be $4x$, then $FG = 3x$ and $CG = x$. Let the perpendicular projection of vertex $C$ onto the base $AB$ be $H$, and introduce the notations $CH = m$ and $\angle ABC = \beta$.

The length of the midline of the trapezoid is $EF = \frac{a+c}{2}$, and due to symmetry, $BH = \frac{a-c}{2}$. The right triangles $EFG$ and $CBH$ are similar because $\angle EFG = \beta$. Therefore, the ratio of the corresponding sides of the two triangles is the same, so $\frac{EF}{FG} = \frac{CB}{BH}$, which means $FG \cdot CB = EF \cdot BH$, from which we get

$$
3x \cdot 4x = \frac{a+c}{2} \cdot \frac{a-c}{2}, \quad \text{or} \quad x^2 = \frac{a^2 - c^2}{48}
$$

In the right triangle $CHB$, by the Pythagorean theorem,

$$
m^2 = (4x)^2 - \left(\frac{a-c}{2}\right)^2
$$

or

$$
m = \sqrt{\frac{a^2 - c^2}{3} - \frac{(a-c)^2}{4}} = \sqrt{\frac{a^2 + 6ac - 7c^2}{12}}
$$

Therefore, the area of the trapezoid is

$$
T = \frac{a+c}{2} \cdot m = \frac{a+c}{2} \cdot \sqrt{\frac{a^2 + 6ac - 7c^2}{12}}.
$$