There are three cases.
Case I: The quadrilateral is a parallelogram. Pick any side and
construct the desired line parallel to it and through the intersection
of the diagonals.
Case II: The quadrilateral is a trapezoid. Label the quadrilateral
ABCD such that AB  CD and point Y is the intersection of rays
AD and BC. Let PQ be the desired line with points P and Q on rays
YA and YB respectively.
Area(ABQP) = Area(PQCD) ⇔
2YPYQ = YAYB + YCYD (1)
PQ  AB ⇔ YAYQ = YBYP (2)
Eliminating YP from (1) and (2) gives
YQ = sqrt( YB*(YB + YCYD/YA)/2 )
EF = YCYD/YA is constructible
GH = YB + EF " "
IJ = GH/2 " "
YQ = sqrt(YBIJ) " "
PQ  AB " "
Case III: No sides of the quadrilateral are parallel. Label the
quadrilateral ABCD such that point Y is the intersection of rays AD
and BC and point X is the intersection of rays AB and DC. Let PQ be
the desired line with points P and Q on rays YA and YB respectively.
The construction is identical to Case II.
The construction fails if YQ < YC. In that case let PQ  AD
be the desired line with points P and Q on rays XA and XD respectively.
Area(ADQP) = Area(PQCB) ⇔
2XPXQ = XAXD + XBXC (3)
PQ  AD ⇔ XAXQ = XDXP (4)
Eliminating XP from (3) and (4) gives
XQ = sqrt( XD*(XD + XBXC/XA)/2 )
EF = XBXC/XA is constructible
GH = XD + EF " "
IJ = GH/2 " "
XQ = sqrt(XDIJ) " "
PQ  AD " "
The construction fails if XQ < XC.
See my "Response to Solution".
QED
