Let ABC be a triangle with P a point on line BC
such that B lies between P and C.

Construct a line through P, intersecting sides
AB and AC in points Q and R respectively, such 
that vertex S of parallelogram QARS lies on 
line BC.

If we were given point S, then the construction would be easy: first draw AS and let its midpoint be I.  Then draw line PI.  The intersections with AB and AC will be the points Q and R and the figure AQSR will be a parallelogram based from the fact that the diagonals of a parallelogram bisect each other, in this case at point I.

Once parallelogram AQSR is constructed, triangles BQS, SRC, and BAC will all be similar.  Since those triangles are similar, then PBQ and PSR are similar and PSQ and PCR are similar.  Furthermore the scale ratios of BQS/SRC, PBQ/PSR, and PSQ/PCR are all the same ratio of BS/SC.

Let PB=x, BS=a, and SC=b.  The problem gives x and a+b as known lengths.  Using the ratios above, PS/PC = PB/PS becomes (x+a)/(x+a+b) = x/(x+a).  This equation rearranges to a=sqrt[x*(x+a+b)]-x.  Fortunately, this expression is constructable.

Draw a line with length 2x+a+b (PB+PC) and place a mark K separating the line into lengths x(=PB) and x+a+b(=PC).  Then draw a circle with the entire segment as the diameter and make a perpendictular line at K.  The distance from K to where the perpendictular meets the circle is x+a, from the equation on the previous paragraph.  Then the length a is taken by shortening the x+a segment by x.

Now with length a known, mark point S on side BC with BS=a.  Then execute the parallelogram construction in the first paragraph to construct the desired parallelogram AQSR.