The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
Inscribe in a given circle a quadrangle in which de two diagonals have given lengths and, in addition, where the ratio between two adjacent sides are equal to a given ratio.
I have no clue on how to solve problems of this kind, possibly since the topic was dropped in the Swedish schools in the mid 1900s and will probably never be taken up again, and I find them very difficult. (Usually the solutions are very elegant and ‘easy’ once you see them but never easy to initially spot for the untrained eye.) The problem should be solved with only a pen, a straightedge and a compass. Problems of this kind were often the final question on the exams, and therefore the hardest one. All help and hints on how to solve this problem are much appreciated.
Initial ideas
Consider the image below where the given circle has been drawn. To the left are the given ratio $r$ and the given two diagonals.
We can place the first diagonal in the circle vertically as we can always rotate the problem around $O$.
We can (I think) remove the quadangle from the problem and rewrite it as: “Set the end points $P$ or $P'$ of the second diagonal so that they are $d_2$ apart and that the distances $PA$ and $P'A$ have the ratio $r$.” There is only need to set one of the end points of the second diagonal as the other end point is easy to draw when the first end point is placed.
The image below shows three such second-diagonal lines. For the 3rd I have drawn the adjacent sides in red. As $P_i$ “slides” on the circle (and $P_i'$ tags along) we ge a ratio from $0$ to $\infty$ (or $0$ to $1$, depending on how you look at the ratio). The question is how to place the second diagonal so it results in the specified ratio. Currently I have no idea.
The original problem
Comment
Not clear in what manner the adjacent sides bear a ratio $p$. Assuming equal sides are placed at opposite positions as
$( a, ap,a, ap),$ then by Ptolemy theorem
$$ a^2(1+p^2) = d_1 d_2 $$
where $ (d_1,d_2) $ are the diagonals...
Using Ruler/Compass a rectangle can be constructed as a special (trivial?) case inside a cyclic quadrilateral.