To construct a symmetrical trapezoid where the length of one of the parallel sides is equal to the length of the legs, given the length of the leg and the distance from the intersection of the diagonals to one endpoint of the other parallel side $(A M=a, A D=b)$.
According to our assumption, the triangle $ADC$ is isosceles, and since the angle at $C$ is the same as the angle at $C$ in the also isosceles triangle $CDM$, the angles $DAM$ and $CDM$ are equal. Therefore, the circle $k$ circumscribed around the triangle $ADM$ is tangent to the line $CD$, since $\angle DAM$ is a peripheral angle subtending the arc $DM$, and $\angle CDM$ is equal to it. Thus, according to the theorem of the segments of a tangent and a secant to a circle,

$$
b^{2}=c(a+c)
$$

where $c=CM$. Therefore, let's draw a circle $k_{0}$ with radius $a/2$, and measure a segment of length $b$ from the tangent at a point $D$ on $k_{0}$, and let the endpoint of this segment be $C$. Connect $C$ to the center of $k_{0}$, and the segment of this line from $C$ to $k_{0}$ is exactly of length $c$, since (1) is still true, and the segment of the secant within $k_{0}$ is exactly the diameter of $k_{0}$. (Since the right-hand side of [1] is a monotonically increasing function of $c$, only one positive $c$ can be a solution.) If we already know $c$, the triangles $CDM$, $ADC$, and the desired trapezoid $ABCD$ can be easily constructed.

![](https://cdn.mathpix.com/cropped/2024_05_02_b61f3627c0aa32fe8670g-1.jpg?height=852&width=1063&top_left_y=669&top_left_x=520)

The construction can be carried out if an isosceles triangle can be constructed with sides $b$ and $c$, i.e., if $2c > b$. Since (1) is equivalent to the equation

$$
(2c - b)(2b + c) = c(3b - 2a)
$$

for positive numbers for which (1) holds, it is true if and only if $3b > 2a$, which is the condition for constructibility. (Intuitively, if $2c$ is close to $b$, then the triangle $CDM$ is very flat, $M$ is close to the midpoint of $CD$, and $AM$ is close to $1.5AD$.)

If the triangle $CDM$ can be constructed, the trapezoid $ABCD$ obtained from it is always suitable, since when we construct a triangle $ABC$ similar to the triangle $CDM$, the length of $AM$ is exactly $a$ due to (1). The problem is therefore always uniquely solvable if $3b > 2a$, otherwise there is no solution.

Note. Equation (1) is the so-called golden section equation. We have already mentioned this equation several times in our journal, for example, in the solution of problem F. 2209 on page 128 of the November 1979 issue. There we mentioned that this construction is nothing other than the division of a segment into two parts according to the golden section, which is a generalization of the Euclidean construction of a regular pentagon. Those interested can read more about this on page 135 of the November 1977 issue of our journal (see also page 201 of the May 1962 issue of KÖMAL).