Three Parallels in a Triangle

The applet attempts to suggest the following statement:

In the proof we shall refer to the following diagram:

As usual BC = a, AC = b, and AB = c. In the diagram a > b > c, which we shall assume but which, in general, may not hold. Thus we consider only one of several possible cases and claim that other cases can be studied by analogy in a similar vein.

We shall only prove that DE and KL are parallel. The claim that MN is parallel to the two can be bundled under the analogy argument above with a redistribution of magnitudes of the lengths a, b, c.

Let KL meet AC in T. According to Menelaus' theorem,

where by construction AK = LC = b, KB = c-b, BL = a - b. (The values of c - b and a - b, as opposed to b - c and b - a, have been selected as to insure agreement with the signs of the terms in (1) implied by Menelaus' theorem.) Denoting CT = x, we rewrite (1) as

from which

One way to show that DE is parallel to KL is to verify the proportion (where all segment length are assumed positive.)

But, with CE = b - c and CD = a - c, (2) becomes

which is indeed true, thus confirming (2).

Nathan Bowler has offered a different and a much simpler approach:

EK is the reflection of BC in the bisector of angle A, so the point X where these lines meet is the intersection of BC with that bisector: It satisfies XB:XC = c:b = BD:CL, so

so DE and KL are parallel.

Michel Cabart suggest a vector algebra shortcut (vectors are in bold):

Subtracting (3) from (4) gives

Multiplying by ab leads to

which is symmetric A, B, C and thus ensures that ab ED = ac LK = bc NM.