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.

Let G be the point on ray CA such that |CG| = c.
The locus of intersections is the line through G
and parallel to the bisector of ∠BAC (excluding
the point where the line intersects line BC).

PROOF:

Let |x| = |BE| = |CD| with x < 0 if points D and E
are on the same side of the line BC as vertex A.
Then

AE = [(c+x)/c]ABAD = [(b+x)/b]AC

AF = (1-λ)AB + λAD

= (1-λ)AB + λ[(b+x)/b]AC (1)

AF = μAE + (1-μ)AC

= μ[(c+x)/c]AB + (1-μ)AC (2)

Solving (1) and (2) for λ and substituting into
(1) gives

AF = [(c+x)AB + (b+x)AC]/(b+c+x)

= [(b-c)/b]AC + [c(c+x)/(b+c+x)](AB/c + AC/b)

= AG + p[AB/c + AC/b]

Therefore, each intersection lies on the line
through G and parallel to the bisector of ∠BAC.

If P is a point on the line through G and parallel
to the bisector of ∠BAC, then

AP = AG + p[AB/c + AC/b]

and

x = [bp - c(c-p)]/(c-p)

will determine points D and E such that P is the
intersection of BD and CE.

Therefore, the claim holds.

QED

Comments: (
You must be logged in to post comments.)