Let's construct a trapezoid that can be inscribed in a given circle, where one of its sides is the diameter of the circle.
Solution. The trapezoid $ABCD$ to be constructed is isosceles, as it is a cyclic trapezoid, and its legs are seen from the endpoints of the longer parallel side at an acute angle, thus they are smaller than the diameter of the circle. Therefore, the side $AB$ of the trapezoid, which is equal to the diameter of the circle, is the longer parallel side.

![](https://cdn.mathpix.com/cropped/2024_05_02_df18ae9e7384d2d46611g-1.jpg?height=523&width=591&top_left_y=308&top_left_x=756)

We determine the length of the leg by calculation. Let $AB=1, AD=BC=x$, and the projection of $D$ onto $AB$ be $G$ (Figure 1). According to Thales' theorem, triangle $ABD$ is a right triangle, so by the known geometric mean theorem, $AD^2 = AB \cdot AG$, which means $AG = x^2$. Therefore, $DC = AB - 2AG = 1 - 2x^2$. Since a circle can be inscribed in the trapezoid, $AD + BC = AB + DC$, which gives

$$
2x = 1 + (1 - 2x^2)
$$

The positive root of this equation is

$$
x = \frac{\sqrt{5}}{2} - \frac{1}{2}
$$

Based on this, the construction is as follows. Through the center $O$ of the given circle, we draw a diameter $AB$ and at the endpoint $B$ of this diameter, we draw the tangent, and measure the segment $BH = BO = 1/2$ on it, then $AH = \sqrt{5}/2$. We intersect the segment $AH$ with the circle centered at $H$ with radius $HB$ at point $J$, then $AJ = x$. Finally, we intersect the circle with the arcs centered at $A$ and $B$ with radius $AJ$ to find the points $D$ and $C$.

Thus, the vertices are on the circle, and (1) is also satisfied, so a circle can be inscribed in the trapezoid, which meets the requirements.

Remark. The above construction can be performed at half scale, by drawing a circle over the radius $OK$ perpendicular to $AB$. Denoting the center of this circle by $L$ and the intersection points of $AL$ by $M$ and $N$ (with $AM < AN$), we have $AM = AJ / 2$, based on which we can find the midpoint $E$ of side $AD$ from the Thales circle with diameter $AO$.

Based on the intersection point $N$, however, we can mark the intersection point $P$ of the legs on the symmetry axis of the trapezoid. Since $OE$, as the altitude of the isosceles triangle $OAD$, is perpendicular to $AP$, from the right triangle $PAO$,

$$
AP = \frac{AO^2}{AE} = \frac{1}{4 \cdot AE} = \frac{1}{4 \cdot AM} = \frac{1}{\sqrt{5} - 1} = \frac{\sqrt{5}}{4} + \frac{1}{4} = AL + LN = AN
$$