Question 1139131
<pre>{{{drawing(400,150,-6,5,-1,8,
locate(-5,0,P), locate(2,8,Q),
line(-5,0,4,0),line(2,7,4,0),line(2,7,-3,5),line(-5,0,-3,5),
locate(-5,0,P), locate(2,8,R), locate(4,0,Q), locate(-3,6,S),
locate(-4,2.5,A),locate(-.5,0,B),locate(3.1,3.9,C),locate(-.5,6,D),


green(line(-4,2.5,-.5,0),line(-.5,0,3,3.5),line(3,3.5,-.5,6),line(-.5,6,-4,2.5))
 )}}}

All you need do is to draw one diagonal, say, PR.

 {{{drawing(400,150,-6,5,-1,8,
locate(-5,0,P), locate(2,8,R), locate(4,0,Q), locate(-3,6,S),
line(-5,0,4,0),line(2,7,4,0),line(2,7,-3,5),line(-5,0,-3,5),

locate(-4,2.5,A),locate(-.5,0,B),locate(3.1,3.9,C),locate(-.5,6,D),

red(line(-5,0,2,7)),
green(line(-4,2.5,-.5,0),line(-.5,0,3,3.5),line(3,3.5,-.5,6),line(-.5,6,-4,2.5))
 )}}}

Then AD is a midline of triangle PRS and thus parallel to and half
the length of PR. 

Similarly, BC is a midline of triangle PRQ and is 
thus ALSO parallel to and half the length of PR.  So AD and BC are both>
parallel and equal in length.  That's enough to prove ABCD is a 
parallelogram.  Now go write that up in a two-column proof.

Edwin</pre>