(In reply to Response to Solution
It seems to me that the construction 'fails' to find an area bisector, PQ,
that crosses two adjacent sides rather than opposite sides – since this
contradicts the initial assumption that P and Q lie on AD and BC.
W.l.o.g. label the quadrilateral so that BA and CD meet when produced,
and AD and BC meet when produced.
Let DE and DF be parallel to AB and CB, with E & F on BC & AB resp.
The quadrilateral is divided into 3 regions:
Parallelogram BEDF (area U), triangle CDE (area V), triangle ADF (area W).
If area bisector ||AB crosses adjacent sides (BC & CD) then V>(U+V+W)/2.
If area bisector ||BC crosses adjacent sides (AB & AD) then W>(U+V+W)/2.
Adding: V + W > U + V + W => U < 0
Clearly not true, so the construction will work in at least one of the two cases.
Edited on March 18, 2014, 1:13 pm
Posted by Harry
on 2014-03-18 13:12:20