mscroggs.co.uk
mscroggs.co.uk

subscribe

Blog

Archive

Show me a random blog post
 2019 
 2018 
 2017 
 2016 
 2015 
 2014 
 2013 
 2012 

Tags

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

Archive

Show me a random blog post
▼ show ▼
 2016-03-31 

Proving Pythagoras' theorem

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 "tcesib" backwards in the box below (case sensitive):
© Matthew Scroggs 2019