Pairs of Homologous Lines under Spiral Similarities

One of the most basic theorems of the theory of three directly similar figures (see [Casey, Johnson, Lachlan, Yaglom]) claims the existence of a circle of similarity of the three figures that contains a variety of special points. The applet below illustrates an extension of this result:

 

Let F1, F2, F3 be any three figures which are directly similar; let O1 be the center of similitude of F2 and F3; O2 that of F3 and F1; and O3 that of Fl and F2. Let there be two triples of homologous lines AkBk and AkCk, k = 1, 2, 3. Three lines AB form a triangle D1D2D3. Three lines AC form a triangle D'1D'2D'3. Both are perspective to ΔO1O2O3 from points on the circle of similitude. In addition, the two triangles are directly similar and can be obtained from one another with a spiral similarity with center on the circle of similitude.

In the applet, two centers of similarity can be dragged and their rotation angles and coefficients modified with the dials on the left side of the applet. ΔA1B1C1 can also be dragged either as a whole or modified by dragging its vertices.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Proof

References

  1. J. Casey, A Sequel to the First Six Books of the Elements of Euclid, University of Michigan, 2005 (reprint of 1888 edition), pp. 189-193
  2. R. A. Johnson, Advanced Euclidean Geometry , Dover, 2007 (reprint of 1929 edition), pp. 302-312
  3. R. Lachlan, An Elementary Treatise on Modern Pure Geometry, Cornell University Library (reprint of 1893 edition), pp. 140-142
  4. I. M. Yaglom, Geometric Transformations II, MAA, 1962, p. 82, pp. 163-165

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Copyright © 1996-2017 Alexander Bogomolny

Proof

That triangles D1D2D3 and D'1D'2D'3 are perspective to ΔO1O2O3 from points on the circle of similitude has been proved elsewhere. We only have to show that they are directly similar and that the center of spiral similarity that maps one on the other is located on the circle of similitude.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Indeed, by the construction, the side lines of triangles D1D2D3 and D'1D'2D'3 are obtained from each other in pairs by fixed rotations. It follows that the corresponding lines in the two triangles are equally inclined to eacher other so that the triangles are similar.

We'll have to consider to separate cases: (a) the triangles have three parallel sides, (b) not all sides are parallel.

... to be continued ...

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