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

 

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Explanation

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Copyright © 1996-2017 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

  AE = BD = AB,
AK = CL = AC,
CM = BN = BC.

Show that that the three lines DE, KL, and MN are parallel.


 

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What if applet does not run?

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,

(1) AK/KB · BL/LC · CT/TA = -1,

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

  (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 CD = a - c, (2) becomes

  (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 XB:XC = c:b = BD:CL, so

  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 ab ED = ac LK = bc NM.

(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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