How to reconstruct a quadrilateral ABCD only using compass and straight edge? Reconstruct a quadrilateral ABCD given length of its sides and the length of the midline between the first and third sides (namely all the segments drawn in the given figure) using compass and straight edge.  The method is parallel translation but i don't know how to do it.

 A: This is really an interesting problem, which appears to be trivial. However, the only solution I can think of is not that trivial:

Let $M, N$ be the midpoints of $AC, BD$ respectively. Then $EMFN$ is a parallelogram, with $EM=BC/2, MF= AD/2$. So $EMFN$ is constructable from the given data. Now let $M_1$ be the reflection of $E$ about $M$ and $G$ the common midpoint of $MN$ and $EF$. It is trivial that $M_1$ is contructable, and that $\vec{EM_1} = \vec{BC}$. Thus $EBCM_1$ is a parallelogram and so $CM_1=EB = AB/2$. Therefore, we can construct $C$ (note that there are two possible solution here). Once we have $C$, 


*

*$D$ is the reflection of $C$ about $F$

*$B$ is the reflection of $D$ about $N$

*$A$ is the reflection of $B$ about $E$
And we are done.
A: Reflect $A,B$ and $E$ across $F$ to $A',B'$ and $E'$. Then $AEA'E'$, $BEB'E'$ and $ABA'B'$ are parallelograms. Also $AEE'B'$ is parallelogram. Let $G$ halves $AB'$, then since $F$ halves $EE'$ we see that $AEFG$ is also parallelogram, so we know the length of $FG$. 
Now the construction:


*

*Draw green triangle $AB'D$ (we know all it sides)

*Draw blue triangle $DGF$ (we know all it sides)

*Draw quadrilateral $ABCD$. 



A: 
During the construction:
Either BC is fixed by compass and then AD is automatically determined,
or,
AD is fixed and then BC is automatically determined.
However BC and AD cannot both be fixed together. 
No construction is possible.
To visualize consider EBCF as a 4-bar mechanism of given link lengths made in steel with EF as fixed link and BC fixed, link AD would break unless made of thin elastic rubber.
