mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

 2016-03-31 
Pythagoras's Theorem is perhaps the most famous theorem in maths. It is also very old, and for over 2500 years mathematicians have been explaining why it is true.
This has led to hundreds of different proofs of the theorem. Many of them were collected in the 1920s in The pythagorean proposition by Elisha Scott Loomis [1]. Let's have a look at some of them.

Using similar triangles

For our first proof, start with a right angled triangle, \(ABC\), with sides of lengths \(a\), \(b\) and \(c\).
Add a point \(D\) on the hypotenuse such that the line \(AD\) is perpendicular to \(BC\). Name the lengths as shown in the second diagram.
\(ABC\) and \(DBA\) are similar triangles, so:
$$\frac{b}{x}=\frac{c}{b}$$ $$b^2=xc$$
\(ABC\) and \(DAC\) are similar triangles, so:
$$\frac{a}{c-x}=\frac{c}{a}$$ $$a^2=c^2-cx$$
Adding the two equations gives:
$$a^2+b^2=c^2$$

Constructing a quadrilateral

This proof shows the theorem is true by using extra lines and points added to the triangle. Start with \(ABC\) as before then add a point \(D\) such that \(AD\) and \(BC\) are perpendicular and of equal length. Add points \(E\) on \(AC\) and \(F\) on \(AB\) (extended) such that \(DE\) and \(AC\) are perpendicular and \(DF\) and \(AB\) are perpendicular.
By similar triangles, it can be seen that \(DF=b\) and \(DE=a\).
As the two diagonals of \(BACD\) are perpendicular, its area is \(\tfrac12c^2\).
The quadrilateral \(BACD\).
The area of \(BACD\) is also equal to the sum of the areas of \(ABD\) and \(ACD\). The area of \(ABD\) is \(\tfrac12b^2\). The area of \(ACD\) is \(\tfrac12a^2\).
The triangles \(ABD\) and \(ACD\).
Therefore, \(\tfrac12a^2+\tfrac12b^2=\tfrac12c^2\), which implies that \(a^2+b^2=c^2\).

Using a circle

This proof again uses extra stuff: this time using a circle. Draw a circle of radius \(c\) centred at \(C\). Extend \(AC\) to \(G\) and \(H\) and extend \(AB\) to \(I\).
By the intersecting chord theorem, \(AH\times AG = AB\times AI\). Using the facts that \(AI=AB\) and \(CH\) and \(CG\) are radii, the following can be obtained from this:
$$(c-a)\times(c+a)=b\times b$$ $$c^2-a^2=b^2$$ $$a^2+b^2=c^2$$

Rearrangement proofs

A popular method of proof is dissecting the smaller squares and rearranging the pieces to make the larger square. In both the following, the pieces are coloured to show which are the same:
Alternatively, the theorem could be proved by making copies of the triangle and moving them around. This proof was presented in The pythagorean proposition simply with the caption "LOOK":

Moving proof

This next proof uses the fact that two parallelograms with the same base and height have the same area: sliding the top side horizontally does not change the area. This allows us to move the smaller squares to fill the large square:

Using vectors

For this proof, start by labelling the sides of the triangle as vectors \(\alpha\), \(\beta\) and \(\gamma\).
Clearly, \(\gamma = \alpha+\beta\). Taking the dot product of each side with itself gives:
$$\gamma\cdot\gamma = \alpha\cdot\alpha+2\alpha\cdot\beta+\beta\cdot\beta$$
\(\alpha\) and \(\beta\) are perpendicular, so \(\alpha\cdot\beta=0\); and dotting a vector with itself gives the size of the vector squared, so:
$$|\gamma|^2=|\alpha|^2+|\beta|^2$$
If you don't like any of these proofs, there are of course many, many more. Why don't you tweet me your favourite.

The pythagorean proposition by Elisha Scott Loomis. 1928. [link]

Similar posts

Harriss and other spirals
World Cup stickers 2018, pt. 3
Mathsteroids
Video game surfaces

Comments

Comments in green were written by me. Comments in blue were not written by me.
 Add a Comment 


I will only use your email address to reply to your comment (if a reply is needed).

Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "rebmun" backwards in the box below (case sensitive):

Archive

Show me a random blog post
 2019 

Sep 2019

A non-converging LaTeX document
TMiP 2019 treasure punt

Jul 2019

Big Internet Math-Off stickers 2019

Jun 2019

Proving a conjecture

Apr 2019

Harriss and other spirals

Mar 2019

realhats

Jan 2019

Christmas (2018) is over
 2018 
▼ show ▼
 2017 
▼ show ▼
 2016 
▼ show ▼
 2015 
▼ show ▼
 2014 
▼ show ▼
 2013 
▼ show ▼
 2012 
▼ show ▼

Tags

plastic ratio cambridge mathslogicbot folding paper geometry national lottery logic talking maths in public statistics interpolation frobel puzzles approximation polynomials golden spiral people maths fractals news data world cup pac-man python chalkdust magazine speed books ternary arithmetic royal baby harriss spiral golden ratio wool triangles electromagnetic field rhombicuboctahedron map projections the aperiodical programming coins reddit light braiding countdown raspberry pi twitter european cup tmip captain scarlet football palindromes machine learning inline code christmas bubble bobble probability dragon curves binary video games sound trigonometry asteroids nine men's morris noughts and crosses game show probability dataset bodmas rugby flexagons error bars pythagoras game of life manchester science festival misleading statistics draughts realhats craft weather station dates graph theory reuleaux polygons gerry anderson accuracy big internet math-off cross stitch folding tube maps javascript sport oeis matt parker stickers curvature latex platonic solids pizza cutting games php hats radio 4 martin gardner london underground mathsteroids manchester sorting mathsjam hexapawn tennis go chess a gamut of games christmas card propositional calculus estimation chebyshev menace london final fantasy

Archive

Show me a random blog post
▼ show ▼
© Matthew Scroggs 2019