
Introduction
In a triangle with rectangle sides of length a en b and hypothenuse c
the equation :
is true.
This equation is known as the Pythagoras theorem.
See picture at right.
This theorem is the basis of many other theorems and calculations.
This article supplies four proofs.
Proof 1
We start with square ABCD.
On the sides, points E,F,G,H are placed such that AE = BF = CG = DH.
We call this distance a.
A side has length a + b.
The proof has two parts:
1. Proof that EFGH is a square.
The construction implies that triangles AEH,BFE,CGF and DHG
are congruent.
So, in any case, EFGH is a rhombus.
The length of a side we call c.
Angles (1) and (3) are 90 degrees together, so angle (2) is 90 degrees.
EFGH is a square.
2. the equation:
Let's write an area as [ ]
Look at the areas and notice that:
[ABCD] = 4 * [AEH] + [EFGH]
Proof 2
Now we start with two unequal squares:
ABCD with a side of a
BEFG with a side of b.
Point H is on AB such, that AH = b.
Which implies that HE = a.
We use the scissors and cut triangles ADH and HEF
placing them on sides DC and GF.
Similar to proof 1 we see, that HFID is a square.
The coloring shows:
Proof 3
First a preamble.
Right are pictured two triangles: ABC and DEF.
These triangles are similar:
One is an enlargement of the other, the multiplication factor is p.
Notice the similar sides: AB,DE, or AC,DF.
Similar triangles have the properties:
 similar angles are equal
 ratio's of similar sides are equal
Conversely: two triangles are similar if their angles are equal.
Lemma
The areas of similar triangles have a ratio equal to the square of ratio's of similar sides
The proof unfolds: (see figure above)
Double the area of the triangles is ab and abp^{2}.
The ratio of areas is 1 : p^{2} which is the ratio of similar sides, power 2.
Picture right has BD perpendicular to AC and AB perpendicular to BC.
There are three similar triangles: ABC, ADB and BDC.
Note the correct sequence of the angles for similarity.
Equal angles are marked equally.
With the previous knowledge we may proceed:
(call AB = a , BC = b , AC = c and notice similar hypothenuses)
We see:
[ADB] + [BDC] = [ABC]
[ADB] : [ABC] + [ABC] : [ABC] = 1
and because of the lemma:
Proof 4
Here we make use of similarity as well.
1.
DABD ~DBCD
c_{1} : b = b : c
c_{1}c = b^{2}
2.
DBCD ~DACB
c_{2} : a = a : c
c_{2}c = a^{2}
1. and 2. added
a^{2} + b^{2} = c_{2}c + c_{1}c = (c_{2} + c_{1})c = c^{2}

