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Topic: Geometry - Equilateral Triangle (Read 1439 times) |
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Earendil
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Geometry - Equilateral Triangle
« on: Apr 16th, 2005, 2:22pm » |
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The three angles of a triangle are called A, B and C. You trissect A using lines a1 and a2, B, using b1 and b2, and C, using c1 and c2. a1 is the closest line to B (from a1 and a2), b1 is the closest line to A (from b1 and b2) and c1 is the closest line to A (from c1 and c2).
a1 and b1 encounter at point D.
a2 and c1 encounter at point E.
b2 and c2 encounter at point F.
Prove that for any triangle with angles A B C, DEF is an equilateral triangle.
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Barukh
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Re: Geometry - Equilateral Triangle
« Reply #1 on: Apr 25th, 2005, 6:01am » |
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As nobody seems to try on this, here’s the history.
The problem is known as Morley’s theorem. It was discovered over a century ago (1899). The following page from Alex Bogomolny’s amazing site provides several different proofs. Interesting reading, strongly recommended.
Besides, there are additional properties of this configuration:
1. Lines AF, BE, CF are concurrent (meet in a point). A geometer would say that triangles ABC and DEF are perspective. This fact was discovered about 40 years ago by P. Yff.
2. If instead of internal angles A, B, C, we trisect their external counterparts, another equilateral triangle is generated.
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