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

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 $\mathcal S_S$.