Through two given points on a circle, construct two parallel chords with a given sum. The problem is from Kiselev's Geometry exercise 317.

Through two given points on a circle, construct two parallel chords with a given sum.

Here is what I have tried so far:
Mark the two points by $A$ and $C$ respectively. If we have constructed such two chords and marked the two other points by $B$ and $D$, the quadrilateral $ABCD$ is an isosceles trapezoid where $AC$ is a diagonal and (without loss of generality) $AB$ and $CD$ are parallel. The midline of the bases measures half of the given sum, and it passes through the midpoint of the diagonal $AC$.
Unfortunately, I could not progress any further from here; I think I should utilize the fact that the 4 points are concyclic and $ABCD$ is an isosceles trapezoid, but I could not find usage of the fact.
Any help would be much appreciated.
 A: Consider the picture:

Let $A$ and $B$ be the two points, and $AC$ and $BD$ be the desired chords; let $AC+BD=a$. Assume first that the situation is like on the picture, i.e. $C$ and $D$ are on the same side with respect to $AB$. Consider central symmetry with respect to $S$, $\mathcal S_S$, where $S$ is the midpoint of $AB$; so $\mathcal S_S(A)=B$ and $\mathcal S_S(B)=A$. Let $\mathcal S_S(C)=C'$ and $\mathcal S_S(D)=D'$. Then $AD'=BD$ and $AD'\parallel BD$, so $A$, $D'$ and $C$ are colinear and $CD'=a$. Also $\angle BCA=\angle BD'A=:\alpha$ as these are inscribed angles over $AB$ in congruent circles. Thus $BC=BD'$.
Note that a triangle congruent to $\triangle BCD'$ can be constructed: $CD'=a$ is known, as well as $\alpha$ (inscribed angle over $AB$ in the given circle). Thus the measure of $BC$ can be constructed, so $C$ can be constructed as well. Finally, $D$ can be trivially constructed.
The case when $C$ and $D$ are on the opposite sides with respect to $AB$ is similar, just one has to consider the translation $\mathcal T_{\overrightarrow{AB}}$ rather than $\mathcal S_S$.
A: 
Let Q, R be the given points. QRBA the given circle. Partition the line segment of summed length QP at A.
Draw a parallel through B and parallelly transfer AP to BR. The point R must lie on the circle because $\alpha,\beta$  are opposite supplementary angles in a cyclic quadrilateral.
Likewise transfer AQ to BS. Draw congruent circle PABS. Let the diameter of circles be $d$. The geometric construction is anti-symmetric with respect to mid-point of AB.
First I approached in the way you suggested.But partitioning into AP, AT instead of AT, AQ led me into errors.
Fwiw found the following relation by Sine Rule relevant to the construction involving a side, diagonal ( of isosceles trapezium AQRB ) and distance between given parallel lines $h$.
$$ r_1\cdot r_2= h\cdot d  $$
