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Pompeiu's theorem

Given an equilateral triangle ABC in the plane, and a point E in the plane of the triangle ABC, the lengths EA, EB, and EC form the sides of a (maybe, degenerate) triangle.

Q.E.D.

Note by Nicolae Sapoval
3 years ago

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Very nice :D Sotiri Komissopoulos · 3 years ago

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I think some description would be more helpful! :) Snehal Shekatkar · 3 years ago

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@Snehal Shekatkar Look for corresponding congruent parts of the red triangle and the original blue segments. Also note that bit in the bottom left that shows an angle of measure 60 degrees. Limao Luo · 3 years ago

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@Limao Luo I know this theorem well enough just as you do.. But those who haven't even heard of this would need some description I think.. Snehal Shekatkar · 3 years ago

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Given \( \Delta ABC\) and the point \(E\). Now, using \(CA\) as base, construct an equilateral \(\Delta ADC\). The position of the point \(E'\) with respect to vertices \(A\),\(D\) and \(C\) of \(\Delta ACD\) is same as the that of point \(E\) with respect to vertices \(A\), \(C\) and \(B\) respectively of \(\Delta ABC\). Let \(\angle EAB = \theta\). Therefore, \(\angle E'AC = \theta\) and \(\angle CAE = 60-\theta\). Hence, \(\angle E'AE = 60^\circ\) . And \(\Delta E'AE\) is an equilateral triangle. Therefore, \(EE' = EA\). Also, \(E'C = EB\) because of our construction. Therefore, we have constructed a degenerate \(\Delta CE'E\) with side lengths \(CE'=EB\) , \(E'E=EA\) and \(EC\) which was to be proved. Mridul Sachdeva · 3 years ago

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@Mridul Sachdeva The only problem with the solution is that it works only for points within the \(\Delta ABC\). A separate proof is required for points outside the triangle. Mridul Sachdeva · 3 years ago

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2 Gulfam Hussain · 3 years ago

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excellent................. Chandra Sekhar · 3 years ago

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Please explain degenerate and Q.E.D and what is the use of this theorem ...... Please i 'm weak at maths Rohitas Bansal · 3 years ago

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@Rohitas Bansal QED is latin "quod erat demonstrandum"; it means 'Which was to be demonstrated'; other forms include "Hence proved", "Which Was What Was Wanted (\(\text{W}^5\))",etc. Sometimes, as a joke "Quite easily done". You can read the full article. Degenerate: A triangle with all points on the same line. Kartikay Kumar · 3 years ago

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What is a 'degenerate' triangle? Kartikay Kumar · 3 years ago

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@Kartikay Kumar A triangle whose vertices are collinear. Jorge Tipe · 3 years ago

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I agree with Snehal. Soham Dibyachintan · 3 years ago

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Can't really understand, can someone please tell me :3 Shashank Sistla · 3 years ago

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@Shashank Sistla Basically he rotated the whole triangle about A by 60 degrees. A lot of length equalities follow and we get a triangle with sides equal in length to EA, EB, and EC. Faraz Masroor · 3 years ago

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q.e.d Vinay Sharma · 3 years ago

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