A right-angled trapezoid's area is equal to half the product of its legs. From which point on the perpendicular leg is the other leg seen at the largest angle?
Using the notations in the figure, the bases of the trapezoid are $AB = a, CD = c$, and the legs are $BC = b, AD = d$. Since it is a right trapezoid, the height of the trapezoid is $m = d$. According to the problem statement,

$$
T_{ABCD} = \frac{bd}{2}
$$

Using the area formula for a trapezoid:

$$
T_{ABCD} = \frac{a+c}{2} \cdot m = \frac{a+c}{2} \cdot d = \frac{bd}{2}
$$

Thus, $a + c = b$.

![](https://cdn.mathpix.com/cropped/2024_05_02_88923ba8b8ad596bb132g-1.jpg?height=458&width=499&top_left_y=606&top_left_x=802)

Let $F$ and $G$ be the midpoints of the legs $b$ and $d$, respectively. Then

$$
BF = FC = \frac{b}{2} = \frac{a+c}{2}
$$

Since $FG$ is the midline, $FG = \frac{a+c}{2}$. This means that the midpoint $G$ of the leg $AD$ lies on the Thales' circle with center $F$ and diameter $BC$. Since the trapezoid is right, the midline is perpendicular to the leg $AD$, making it a tangent to the Thales' circle at point $G$.

The point $G$ sees the other leg at a right angle, while any other point on the leg $AD$ lies outside the circle, thus seeing the other leg at an angle less than $90^\circ$.

Therefore, the point we are looking for is the midpoint of the perpendicular leg, the point $G$.