Given a circle $\Gamma$ and the positive numbers $h$ and $m$,  construct with straight edge and compass a trapezoid inscribed in $\Gamma$, such that it has altitude  $h$ and the sum of its parallel sides is $m$.
1. **Assume the Trapezoid and Define Variables:**
   Suppose 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 \(C\) onto the line \(AB\). Then \(AC' = \frac{AB}{2} + \frac{CD}{2} = k\).

2. **Right Triangle and Trigonometric Relationship:**
   From the right triangle \(AC'C\), we have:
   \[
   \tan \theta = \frac{CC'}{AC'} = \frac{h}{k}
   \]

3. **Construction Steps:**
   - Let \(\Gamma = (O, R)\) be the given circle.
   - Construct a right triangle with two perpendicular sides \(h\) and \(k\) and hypotenuse \(w\). Then the angle between \(k\) and \(w\) is \(\theta\).
   - Construct an arc \(BC\) on the circle such that \(\frac{BC}{2R} = \frac{h}{w}\). In other words, \(BC = 2R \cdot \sin \theta\).
   - Construct a triangle \(BC'C\), right at \(C'\), with \(CC' = h\).
   - Extend \(BC'\) to meet the circle at \(A\).
   - Through \(C\), draw a parallel to \(AB\) which meets the circle at \(D\).

4. **Proof of Construction:**
   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 \(\tan (\angle BAC) = \frac{h}{k}\). Therefore, we have \(AB + CD = 2k\).

5. **Conditions for Validity:**
   - **First Condition:**
     \[
     h < 2R \cdot \sin \theta \iff h < 2R \cdot \frac{h}{w} \iff \boxed{w < 2R}
     \]
   - **Second Condition:**
     The quadrilateral \(ABCD\) must be convex. So, the points \(A\) and \(D\) must be on the same side of \(BC\). If \(d\) is the perpendicular bisector of \(AB\), then \(C\) must be in the same half-plane as \(B\). Therefore:
     \[
     AC > BC \iff \angle ABC > \angle BAC \iff \sin(\angle ABC) > \sin(\angle BAC) \iff \frac{CC'}{BC} > \sin \theta \iff \frac{h}{BC} > \sin \theta \iff h > 2R \cdot \sin^2 \theta \iff h > 2R \frac{h^2}{w^2} \iff \boxed{w^2 > 2Rh}
     \]