Let XY denote the vector from point X to point Y.

Let P be a point in the plane of triangle ABC.

Let PQ = PA + PB + PC.

What is the locus of points Q as P traces the triangle ABC ?

Given PQ=PA+PB+PC
Case(i):
Let P be the point on BC
then PQ-PA=AQ, let PC=x*BC(Where x is a scalar)
So, PB=CB(1-x)=BC(x-1)
=> AQ=PB+PC
=> AQ=BC(2x-1)
From the above AQ segment is parallel to BC
when x=1 AQ=BC and when x=0, AQ=CB
So, when P is on BC segment, locus of Q is a line segment
parallel to BC of twice the length of BC and whose 
midpoint is A.
Similarly the same with P on other segments

So, locus of Q is a triangle, with mid points of sides
A,B and C and twice as lengths of AB,BC and CA.

Posted by Praneeth on 2008-05-16 03:40:34