Fermat Points and Concurrent Euler Lines
Let F1 and F2 denote the (inner and outer) Fermat-Toricelli points of a given ΔABC. We prove that the Euler lines of the 10 triangles with vertices chosen from A, B, C, F1, F2 (three at a time) are concurrent at the centroid of ΔABC [Beluhov]. This is obviously true for ΔABC itself. The applet below illustrates the case where the triangles are formed by two Fermat's points and one of the vertices A, B, C. There are three such triangles. The other triangles are considered separately.
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Copyright © 1996-2015 Alexander Bogomolny
In the applet,
- A'B'C' is the outer and a'b'c' the inner Napoleon's triangles,
- F and f are the corresponding Fermat's points,
- G and G' are the centroids of ΔABC and ΔFfC, respectively,
- O is the circumcenter of ΔFfC,
- P the intersection of B'C' with a'c',
- M is the midpoint of AB, m the midpoint of Ff,
- F' the reflection of f in M.
- Napoleon's Theorem
- A proof with complex numbers
- A second proof with complex numbers
- A third proof with complex numbers
- Napoleon's Theorem, Two Simple Proofs
- Napoleon's Theorem via Inscribed Angles
- A Generalization
- Douglas' Generalization
- Napoleon's Propeller
- Napoleon's Theorem by Plane Tessellation
- Fermat point
- Kiepert's theorem
- Lean Napoleon's Triangles
- Napoleon's Theorem by Transformation
- Napoleon's Theorem via Two Rotations
- Napoleon on Hinges
- Napoleon on Hinges in GeoGebra
- Napoleon's Relatives
- Napoleon-Barlotti Theorem
- Some Properties of Napoleon's Configuration
- Fermat Points and Concurrent Euler Lines I
- Fermat Points and Concurrent Euler Lines
- Escher's Theorem
- Circle Chains on Napoleon Triangles
- Napoleon's Theorem by Vectors and Trigonometry
- An Extra Triple of Equilateral Triangles for Napoleon
- Joined Common Chords of Napoleon's Circumcircles
- Napoleon's Hexagon
- Fermat's Hexagon
- Lighthouse at Fermat Points
- Midpoint Reciprocity in Napoleon's Configuration
We only prove the claim for ΔFfC.
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What if applet does not run?As we know, the two Napoleon triangles share the centroid G. Both are invariant under the rotations through 120° with center G. One such rotation maps A'B' and a'b' onto B'C' and c'a', respectively. Therefore, it also maps O to P, implying that
∠OGP = 120°. Since∠Oa'P = 120°, the four points O, G, a', P are concyclic. Their circumcircle also contains B' because∠PB'O = 60°. Therefore,(1)
∠OGB' = ∠Oa'B'.
The same rotation maps ∠Oa'B' onto ∠Pc'C', showing that the two are equal:
(2)
∠Oa'B' = ∠Pc'C'.
Since Pc'⊥Bf and C'c'⊥ BA,
(3)
∠Pc'C' = ∠fBA.
Further, since ∠BF'A = ∠AFB = 120° = 180° - ∠AfB, the quadrilateral AfBF' is cyclic, giving ∠fBA = ∠fF'A. It follows that
(4)
∠fF'A = ∠OGB'.
Now, from AF' || fB⊥ A'C' and B'G⊥ A'C', we see that
AF' || B'G. This, together with (4), yields(5)
F'f || GO.
Points G and G' divide the segments CM and Cm in ratio 2:1, causing
GG' || Mm. By construction,FM = MF' and, since alsoFm = mf, the same argument shows thatMm || F'f. Combining this with (5) shows thatGG' || GO, implying the collinearity of G', G and O.References
Napoleon's Theorem
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Copyright © 1996-2015 Alexander Bogomolny
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