Three Parallels in a Triangle: What Is It About?
A Mathematical Droodle

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Copyright © 19962017 Alexander Bogomolny
Three Parallels in a Triangle
The applet attempts to suggest the following statement:
On the sides of ΔABC six points D, E, K, L, M, N are constructed (see the applet) so that
Show that that the three lines DE, KL, and MN are parallel. 

In the proof we shall refer to the following diagram:
As usual BC = a, AC = b, and AB = c. In the diagram
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,
(1)  AK/KB · BL/LC · CT/TA = 1, 
where by construction
(a  b) / (c  b) = (x  b) / x = 1  b/x, 
from which
x = b(b  c) / (a  c). 
One way to show that DE is parallel to KL is to verify the proportion (where all segment length are assumed positive.)
(2)  CE/CT = CD/CL. 
But, with CE = b  c and
(b  c)(a  c) / b(b  c) = (a  c)/b, 
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
XD:XL = (BD  XB):(CL  XC) = c:b = XE:XK, 
so DE and KL are parallel.
Michel Cabart suggest a vector algebra shortcut (vectors are in bold):
(3)  CE  = CA  c/b CA 
(4)  CD  = CB  c/a CB 
Subtracting (3) from (4) gives
ED  = AB  c/a CB + c/b CA 
Multiplying by ab leads to
ab ED  = ab AB + bc BC + ca CA 
which is symmetric A, B, C and thus ensures that
(This statement has a bearing on an interesting property of the line joining the incenter and circumcenter of a triangle. In addition, the three lines are parallel to the line incident with the points of intersection of the external angle bisectors with the opposite sides. Also, points D, E, K, L, M, N are split into pairs lying on each of the side lines of ΔABC. Calculations show that the points in a pair are equidistant from the point of tangency of the incircle of ΔABC with the corresponding side line.)
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Copyright © 19962017 Alexander Bogomolny
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