Morley's Miracle |
| The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle. |
Proof
This proof was published in The American Mathematical Monthly, Vol. 50, No. 9 (Nov., 1943), p. 552.
Let the triangle have base BC and angles 3α, 3β,3γ. Let BDK, BF, CDH, CE be angle trisectors. E is determined by making
| ∠EDF = 360° - (180° - β - γ) - (60° + β) - (60° + γ) = 60°. |
Also
| ∠BFD = 180° - (60° + β + γ) = 60° + α. |
Similarly, ∠CED = 60° + α.
|
Since D is equidistant from BF and CE,
| ∠1 = (60° + α) - (β - γ) = 60° - β. |
Similarly,
| ∠2 = 60° - γ. |
Through F draw line r making
| ∠3 = (60° + α) - (60° - β) = α + β |
and
| ∠mr = (α + β) - β = α. |
Similarly,
| ∠sn = (α + γ) - γ = α. |
Further,
| ∠mn = (180° - 3β - 3γ) = 3α. |
It remains only to prove that the lines m, n, r, and s converge to a point. The line KF joins the vertices of two isosceles triangles and therefore bisects ∠K. Then in triangle mBKs the bisector of ∠ms passes through F and being parallel to r, coincides with it. Similarly in triangle rHCn the bisector of ∠rn passes through E and being parallel to s, coincides with it.
Morley's Miracle
- J.Conway's proof
- D. J. Newman's proof
- Bankoff's proof
- B. Bollobás' proof
- Another proof
- Nikos Dergiades' proof
- G. Zsolt Kiss' proof
- M. T. Naraniengar's proof
- Doodling and Miracles
- Morley's Pursuit of Incidence
- Lines, Circles and Beyond
- On Motivation and Understanding
- Bankoff's conundrum
- Of Looking and Seeing
- Morley's Redux and More, Alain Connes' proof
- An Unexpected Variant
- Proof by B. Stonebridge and B. Millar
- Proof by B. Stonebridge
- Proof by Nolan L Aljaddou
- Proof by Roger Smyth
- Proof by H. D. Grossman
- Proof by R. J. Webster
- Proof by H. Shutrick
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Copyright © 1996-2012 Alexander Bogomolny
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