Let ABC be a triangle with points D and E lying on lines AC and AB respectively such that D and E are on the same side of line BC and |BE| = |CD| > 0. Let F be the intersection of rays BD and CE.

What is the locus of the intersections F?

Prove it.

Taking the vertex, O, of the parallelogram ABOC as the origin, and denoting the position vectors of A, B... , by a, b ..., with magnitudes a, b..., it follows that:

            e = b + (x/c)c   and   d = c + (x/b)b

where x is the (directed) distance of E from B and D from C.

If p and q denote CF as a proportion of CE and BF as a proportion of BD respectively, then
            f = pe + (1 – p)c           =  p[b + (x/c)c] + (1 – p)c         (1)
and also:          
            f = qd + (1 – q)b           =  q[c + (x/b)b] + (1 - q)b

Equating coefficients of b and c in these two expressions for f:

            p = qx/b  +  1  -  q         and       q = px/c  +  1  -  p

Solving simultaneously, these give           p = (x/b)/(x/c + x/b – x²/bc)

                                                            p = c/(b + c – x)            (x not 0)
Substituting in (1):

                   f    =  (c/(b + c – x))b + [(c/(b + c – x))(x/c – 1) + 1]c
                       
                        = (1/(b + c – x))(cb + bc)

Since |cb| = |bc| = bc, it follows that f bisects the angle BOC and its magnitude depends on x. So, in relation to the original triangle ABC, F lies on a line parallel to the bisector of angle A and passes through the opposite vertex of the parallelogram ABOC.
When x = b + c (i.e. BE = AC + AB), there is a singularity giving F as a point at infinity. For x > b + c, f changes sign so that F describes the part of the locus on the other side of O.

Posted by Harry on 2012-03-13 13:58:28