Suppose that we have constructed the isosceles trapezoid $ABCD$ with $AB \parallel CD$ Let $\theta = \angle CAB$ and $k=\frac{m}{2}$ Let $C'$ be the projection of $C$ to the line $AB$. Then $AC' = \frac{AB}{2}+\frac{CD}{2}= k$ From the right triangle $AC'C$ we have $\tan\theta = \frac{CC'}{AC'}= \frac{h}{k}$ Construction Let $\Gamma = (O,R)$ be the given circle We construct a right triangle with two perpendicular sides $h$ and $k$ and hypotenuse $w$. Then the angle between $k$ and $w$ is $\theta$ We construct an arc $BC$ on the circle such that $\frac{BC}{2R}= \frac{h}{w}$. In other words it is $BC=2R\cdot\sin\theta$ We construct a triangle $BC'C$, right at $C'$, with $CC'=h \mbox{ }^{\star(1)}$ We extend $BC'$ to meet the circle at $A$ Through $C$ we bring a parallel to $AB$ which meets the circle at $D\mbox{ }^{\star(2)}$ Proof If the above constructed quadrilateral $ABCD$ is convex, then it is a trapezoid with altitude $h$, it is inscribed in $\Gamma$ and $\angle BAC$ satisfies the property $\tan (\angle BAC) = \frac{h}{k}$, so we have $AB+CD=2k$ Discussion $(1)$st condition $h<2R\cdot \sin\theta \iff h<2R\cdot \frac{h}{w}\iff \boxed{w<2R}$ $(2)$nd condition The quadrilateral $ABCD$ must be convex. So, the points $A,D$ must be on the same side of $BC$ So, if $d$ is the perpendicular bisector of $AB$ then $C$ have to be in the same half-plane with $B$ So $AC>BC\iff$ $\angle{ABC}> \angle{BAC}\iff$ $\sin(\angle ABC) > \sin(\angle BAC) \iff$ $\frac{CC'}{BC}> \sin\theta \iff$ $\frac{h}{BC}> \sin\theta \iff$ $h>2R\cdot\sin^{2}\theta \iff$ $h>2R\frac{h^{2}}{w^{2}}\iff$ $\boxed{w^{2}>2Rh}$ Note Another way to construct the point $A$, after finding the arc $BC$ is to use the fact that $BA=w$

Let $ K,L,M,N$ be the midpoints of $ AB,BC,CD,DA.$ It is clear that $ ABCD$ is an isosceles trapezoid $ \Longrightarrow$ $ KLMN$ is a rhombus with known diagonals $ KM = h$ and $ LN = \frac {_1}{^2}(AB + CD) = \frac {_1}{^2}m.$ The construction of a rhombus congruent to $ KLMN$ is immediate and then we obtain the measure of the diagonal $ AC$ which is twice the length of its side. Triangle $ \triangle ACD$ with side length $ AC = 2KL,$ altitude $ h$ onto $ DC$ and circumcircle $ \Gamma$ is constructible, this is: Fix the chord $ AC$ in $ \Gamma$ and draw the circumference with diameter $ AC.$ Circumference centered at $ A$ with radius $ h$ cuts the circumference with diameter $ AC$ at the orthogonal projection $ H$ of $ A$ onto $ DC,$ then ray $ CH$ cuts $ \Gamma$ at $ D.$ The parallel line to $ CD$ passing through $ A$ cuts $ \Gamma$ again at $ B,$ which completes the trapezoid.