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.

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.

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## Comments

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TopNewestVery nice :D – Sotiri Komissopoulos · 2 years, 10 months ago

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I think some description would be more helpful! :) – Snehal Shekatkar · 2 years, 10 months ago

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– Limao Luo · 2 years, 10 months ago

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.Log in to reply

– Snehal Shekatkar · 2 years, 10 months ago

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..Log in to reply

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 · 2 years, 10 months ago

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– Mridul Sachdeva · 2 years, 10 months ago

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.Log in to reply

2 – Gulfam Hussain · 2 years, 10 months ago

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excellent................. – Chandra Sekhar · 2 years, 10 months 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 · 2 years, 10 months ago

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full article. Degenerate: A triangle with all points on the same line. – Kartikay Kumar · 2 years, 10 months ago

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 theLog in to reply

What is a 'degenerate' triangle? – Kartikay Kumar · 2 years, 10 months ago

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– Jorge Tipe · 2 years, 10 months ago

A triangle whose vertices are collinear.Log in to reply

I agree with Snehal. – Soham Dibyachintan · 2 years, 10 months ago

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Can't really understand, can someone please tell me :3 – Shashank Sistla · 2 years, 10 months ago

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– Faraz Masroor · 2 years, 10 months ago

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.Log in to reply

q.e.d – Vinay Sharma · 2 years, 10 months ago

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