Bottema with Similar Triangles

What Might This Be About?

27 May 2014, Created with GeoGebra

Problem

Assume triangles $BET,$ $DAT,$ and $DFC$ are similar as shown below.

Bottema with similar triangles

Prove that $K,$ the midpoint of $EF$ is independent of the position of $D.$

Solution

Bottema with similar triangles

Assuming the point names as complex numbers, the similarity of the three triangles could be expressed as

$\displaystyle\frac{T-E}{B-E}=\frac{T-A}{D-A}=\frac{C-F}{D-F}=u,$

for a complex number $u.$ From here,

$\displaystyle\frac{(T+C)-(E+F)}{(B+D)-(E+F)}=u,$

implying

$\displaystyle\frac{(A+C)-(E+F)}{(A+B)-(E+F)}=u,$

which exactly means that $E+F$ and, with it, $\displaystyle\frac{E+F}{2}$ is independent of $D.$

Acknowledgment

This problem is a follow-up of the one posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page. The simple solution above has been posted by Leo Giugiuc (Romania) who also observed that Dao's problem extends to the one above. Here's Dao's formulation:

Triangles $BET,$ $TAD,$ $DFC$ are similar isosceles. Prove that the midpoint $K$ of $EF$ is independent of the position of $D.$

General Bottema's theorem with extra rotations - Dao's formulation with isosceles triangles

Bottema's Theorem

  1. Bottema's Theorem
  2. An Elementary Proof of Bottema's Theorem
  3. Bottema's Theorem - Proof Without Words
  4. On Bottema's Shoulders
  5. On Bottema's Shoulders II
  6. On Bottema's Shoulders with a Ladder
  7. Friendly Kiepert's Perspectors
  8. Bottema Shatters Japan's Seclusion
  9. Rotations in Disguise
  10. Four Hinged Squares
  11. Four Hinged Squares, Solution with Complex Numbers
  12. Pythagoras' from Bottema's
  13. A Degenerate Case of Bottema's Configuration
  14. Properties of Flank Triangles
  15. Analytic Proof of Bottema's Theorem
  16. Yet Another Generalization of Bottema's Theorem
  17. Bottema with a Product of Rotations
  18. Bottema with Similar Triangles
  19. Bottema in Three Rotations
  20. Bottema's Point Sibling

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