Maths 1 Extension Notes 3
Not Examinable
Pythagoras’ Theorem
In any right-angled triangle, the square of the length of the
hypotenuse is equal to the sum of the squares of the lengths of
the other two sides.
a2 + b2 = c2
Proof. Consider a square with side lengths equal to a + b.
We can use its edges to make four right-angled triangles:
Since each triangle is right-angled with side lengths a and b, we know that all the hypotenuses
have the same length c.
1
When we remove these triangles we are left with a quadrilateral whose sides all have length c.
The diagram below proves that the interior angle θ of this quadrilateral is a right-angle. By
symmetry, all the interior angles are right-angles.
Thus the quadrilateral is a square with area c2 .
Since
Area of original square − 4× area
= area of quadrilateral,
We have
(a + b)2 − 4 × area
But the area of a triangle of width a and height b is
= c2 .
1
ab. Hence
2
1
(a + b)2 − 4 × ab = c2
2
a2 + 2ab + b2 − 2ab = c2
a2 + b2 = c2 .
2
Another proof of Pythagoras’ Theorem
Another proof of Pythagoras’ Theorem consists of adding squares to the edges of the triangle
as follows:
Before we prove the theorem, we need the following lemma:
Lemma 1. Consider any rectangle
and add a right angled triangle as follows:
where AX can be any length.
Then by drawing the line segment AZ:
we form a triangle 4AY Z such that
Area(2W XY Z) = 2 × Area(4AY Z).
Proof. Add a point X 0 to construct a rectangle 2XAX 0 Y , as follows
3
Then
Area(2W XY Z) =
=
=
=
=
Area(2W AX 0 Z) − Area(2XAX 0 Y )
2 × Area(4ZAX 0 ) − 2 × Area(4AX 0 Y )
2 × (Area(4AY Z) + Area(4AX 0 Y )) − 2 × Area(4AX 0 Y )
2 × Area(4AY Z) + 2 × Area(4AX 0 Y ) − 2 × Area(4AX 0 Y )
2 × Area(4AY Z), as required.
We now start the proof of the theorem by adding three squares, and three line segments, to
the original triangle as follows:
Now note that
d = 90◦ + CBA
d = CBG.
d
DBA
It follows that 4DBA = 4CBG. Now, by using Lemma 1, we have
Area(2DEOB) = 2 × Area(4DBA)
= 2 × Area(4CBG)
= Area(2AHGB).
We now use a similar argument to show that Area(2EF CO) = Area(2CJIA).
This time start with:
4
Note that
d = JCB.
d
Fd
CA = 90◦ + OCA
It follows that 4F CA = 4JCB. Now, by using Lemma 1, we have
Area(2EF CO) = 2 × Area(4F CA)
= 2 × Area(4JCB)
= Area(2CJIA).
This completes the proof, since we have shown that the area of the bottom square is the
sum of the areas of the other two squares.
5
References
1. “College Geometry” by David Kay, pages 4–5, and page 173. Education Resource
Centre library (Melbourne University): 516 KAY 1.
2. “Geometry” by Ken Seydel, page 163. Education Resource Centre library (Melbourne
University): 516 SEY 1A.
3. “The Pythagorean Proposition” by E. S. Loomis. Education Resource Centre library
(Melbourne University): 516.2 LOO 1. (This book has over 200 proofs of Pythagoras’
Theorem!)
6