DEFINITION

Let [XYZ] denote the signed area of ΔXYZ ( positive
if the directed angle XYZ is counterclockwise ).

CONSIDER THE FOLLOWING

Let point P be in the plane of ABCD ( a convex quad-
rilateral that is not a parallelogram ). What is the
locus of of points P such that

    [APB] + [CPD] = [BPC] + [DPA] ?

Prove it.

Let a, b, c, d be the position vectors of A, B, C, D with respect to the origin P.
Consider the vector product:
     (a + c) x (b + d)       =  a x b  +  a x d  +  c x b  +  c x d
                                    =  a x b  +  c x d   -  b x c   -  d x a
                                    =  2([APB]  +  [CPD]  -  [BPC]  -  [DPA])n      (1)

where n is a unit vector perpendicular to the plane containing all the points.

If [APB] + [CPD] = [BPC] + [DPA], it follows that the RHS of (1) is 0 and therefore that (a + c) x (b + d) = 0.

Since  (a + c)/2 and (b + d)/2 are the position vectors of the mid points of AC and BD, this means that these mid points are collinear with P.
So the locus of P is the line passing through the mid points of AC and BD

Posted by Harry on 2012-05-08 13:00:49