# Three Similar Triangles

What is that about?

A Mathematical Droodle

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Copyright © 1996-2018 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.

Think of the vertices as complex numbers. Two triangles ABC and A'B'C' are similar iff, say,

(1) |

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

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

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 *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, λ_{1}, and μ_{1} with _{1} + μ_{1} = 1._{1}_{1}

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

(4) | (λ - λ_{1})C + (μ - μ_{1})C' = 0 |

and from the assumprions _{1} + μ_{1} = 1

(5) | (λ - λ_{1}) + (μ - μ_{1}) = 0. |

If λ ≠ λ_{1}, (4) and (5) imply C = C' contrary to our assumption.

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

### References

- D. Wells,
*You Are a Mathematician*, Dover, 1970

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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