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.

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.

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.

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.

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.

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.

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## Comments

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TopNewestVery nice :D

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I think some description would be more helpful! :)

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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.

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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..

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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.

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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.

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q.e.d

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Can't really understand, can someone please tell me :3

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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.

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I agree with Snehal.

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What is a 'degenerate' triangle?

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A triangle whose vertices are collinear.

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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

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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.

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excellent.................

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2

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