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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestVery nice :D

Log in to reply

I think some description would be more helpful! :)

Log in to reply

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

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.

Log in to reply

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

q.e.d

Log in to reply

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

Log in to reply

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

I agree with Snehal.

Log in to reply

What is a 'degenerate' triangle?

Log in to reply

A triangle whose vertices are collinear.

Log in to reply

Please explain degenerate and Q.E.D and what is the use of this theorem ...... Please i 'm weak at maths

Log in to reply

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.

Log in to reply

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

Log in to reply

2

Log in to reply