Midline in Similar Triangles: What is this about?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet is intended to suggest the following proposition (Ayoub B. Ayoub, Problem 1589, Math Magazine,v 74, n 1 (Feb 2001)):

On the sides of ΔABC, three similar triangles, AKB, BLC, and CNA, are drawn outward. If AB and KL are bisected by D and E, respectively, prove that DE is parallel to NC and determine DE/NC.


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A trigonometric solution has been published in Math Magazine, v 74, n 1, February 2001. Several solvers have also observed that the triangles can be formed inwardly as well as outwardly.

Another solution is suggested by the observation that the problem is highly asymmetric. If F is the midpoint of BC, and the problem is correct, then EF outght to be parallel to AN. Since DF is the midline in ΔABC, DF||AC and DF = AC/2, this would imply (and be implied by the fact) that two triangles DEF and CNA are similar with coefficient 1/2.

But the similarity of these triangles is an immediate consequence of the Three Similar Triangles Theorem.

Prof. W. McWorter offered an algebraic solution. Since the three attached triangles are similar and similarly oriented, K is obtained from B by a spiral similarity with center at A:

K = A + aT(B - A),

where a > 0 is real and T is a fixed rotation. Similarly, L = B + aT(C - B), and N = C + aT(A - C). (Since T can be either a clockwise or counterclockwise rotation, the triangles can be formed internally as well as externally.)

Thus, the vector N - C = aT(A - C). E is half of

L + K = B + aT(C - B) + A + aT(B - A)= (1 - aT)A + B + aTC,

and D is half of A + B. So, D - E is half of

A + B - [(1 - aT)A + B + aTC] = aT(A - C) = N - C.

Hence DE is parallel to CN and half as long.


Related material
Read more...

  • Thales' Theorem
  • Midline in Triangle
  • Midline in Trapezoid
  • Bimedians in a Quadrilateral
  • Midline in Similar Triangles

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