$ABCD -$ is a trapezoid. $AD||BC, AB=CD$. The diagonals of the $AC$ and $BD$ intersect at the point $P$, and the straight lines $AB$ and $CD$ intersect at the point $Q$. Points $O_1$ and $O_2$ are the centers of circles described around triangles $ABP$ and $CDP$, and $r$ is the radius of these circles.
Construct the trapezoid $ABCD$ with straightedge and compass, given $O_1O_2,PQ$ and $r$.
How many solutions for this problem?
I construct (red lines):
1) $\triangle {O_1O_2P} \left(O_1O_2, O_1P=O_2P=r\right)$
2) circles $\omega_1(O_1;r)$ and $\omega_2(O_2;r)$
3) $QP \perp O_1O_2$
I need help here
Lines $BD$ and $AC$ meet at $P$ if $\angle BPA + 2\angle APP'=\pi$, where $P'$ is the other intersection point of the circles. Let $H$ be the midpoint of $AB$: by well-known theorems we have $\angle HO_1A=\angle BPA$ and $\angle AO_1P'=2\angle APP'$.
It follows that the above condition is the same as $\angle HO_1P'=\pi$, in other words: trapezoid diagonals meet at $P$ if $P'O_1$ and $AB$ are perpendicular. To construct the trapezoid you just have to draw the lines through $Q$ perpendicular to $P'O_1$ and $P'O_2$.
Notice that this construction does not work for any value of the data: in some cases those perpendiculars don't intersect the circle and the trapezoid doesn't exist.