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.

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)candd = 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-qandq = px/c+1-p

Solving simultaneously, these give p = (x/b)/(x/c + x/b – x^{2}/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.