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Three Parallels in a Triangle: What Is It About?
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Copyright © 1996-2008 Alexander Bogomolny
Three Parallels in a Triangle
The applet attempts to suggest the following statement:
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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. |
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In the proof we shall refer to the following diagram:
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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
| (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 Boweler 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.
(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 exteranl 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.)
Copyright © 1996-2008 Alexander Bogomolny
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