Three Similar Triangles
What is that about?
A Mathematical Droodle

1 June 2015, Created with GeoGebra

Explanation

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Copyright © 1996-2017 Alexander Bogomolny

Explanation

This is what the applet purports to show.

Take any two similar triangles ABC and A'B'C' whose vertices are traversed in the same direction, either both clockwise or both counterclockwise. Divide AA', BB', CC' in equal proportions and connect the resulting points. The new triangle will be similar the the given two.

Theorem of directly similar triangles

Think of the vertices as complex numbers. Two triangles ABC and A'B'C' are similar iff, say, (B - A)/(C - A) = (B' - A')/(C' - A'). The condition means that not only the ratio of lengths of the pairs of sides AB/AC and A'B'/A'C' are equal but that also the angles between them are the same. With the use of determinants the condition for (direct) similarity of two triangles is simply

(1)

The assertion then follows from the well known properties of the determinants:

  1. If a row or a column is multiplied by a constant factor, then the determinant is multiplied by the same factor.

  2. A determinant does not change if a column (row) is added to another column (row.)

(2)

The first identity implies the second for any λ and μ, not both 0. In particular, (1) implies (2) for λ + μ = 1. The latter is a particular case of the Fundamental Theorem of Directly Similar Figures: if the lines connecting the corresponding vertices of two directly similar polygons are devided in equal ratios, then the resulting polygon is directly similar to the given two.

We may also assert a partial converse. Assume (1) and

(3)

for some λ, μ, λ1, and μ1 with λ + μ = 1 and λ1 + μ1 = 1. Assume also that C is different from C' and that the triangles are not degenerate. Then λ = λ1 and μ = μ1.

Indeed, from (1) and (3) we derive

(4) (λ - λ1)C + (μ - μ1)C' = 0

and from the assumprions λ + μ = 1 and λ1 + μ1 = 1 it follows that

(5) (λ - λ1) + (μ - μ1) = 0.

If λ ≠ λ1, (4) and (5) imply C = C' contrary to our assumption.

The theorem just proved happens to be very useful for solving other problems:

  1. Parallelogram and Similar Triangles
  2. Midline in Similar Triangles

References

  1. D. Wells, You Are a Mathematician, Dover, 1970

Related material
Read more...

  • Angle Bisectors In Rectangle
  • Segment Trisection Induced by Parallels to Medians
  • The Nature of Pi
  • Parallelogram and Similar Triangles
  • Two Triples of Similar Triangles
  • Similar Triangles on Sides and Diagonals of a Quadrilateral
  • Point on Bisector in Right Angle
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