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.

#CosinesGroup #ProofWithoutWords

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Very nice :D

Sotiri Komissopoulos - 7 years, 5 months ago

I think some description would be more helpful! :)

Snehal Shekatkar - 7 years, 5 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.

Limao Luo - 7 years, 5 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..

Snehal Shekatkar - 7 years, 5 months ago

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 - 7 years, 5 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.

Mridul Sachdeva - 7 years, 5 months ago

q.e.d

Vinay Sharma - 7 years, 5 months ago

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

Shashank Sistla - 7 years, 5 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.

faraz masroor - 7 years, 5 months ago

I agree with Snehal.

Soham Dibyachintan - 7 years, 5 months ago

What is a 'degenerate' triangle?

Kartikay Kumar - 7 years, 5 months ago

A triangle whose vertices are collinear.

Jorge Tipe - 7 years, 5 months ago

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

Rohitas Bansal - 7 years, 5 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 the full article. Degenerate: A triangle with all points on the same line.

Kartikay Kumar - 7 years, 5 months ago

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

Chandra Sekhar - 7 years, 5 months ago

Gulfam Hussain - 7 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$