Pythagoras' theorem — that the square on the hypotenuse equals the sum of the squares on the other two sides — can be proved without measuring anything. The most elegant proof uses four identical right-angled triangles rearranged inside a square to show that c² = a² + b² must be true for every right-angled triangle.
What is Pythagoras' theorem?
For any right-angled triangle with legs of length a and b and hypotenuse of length c:
a² + b² = c²
The hypotenuse is the side opposite the right angle and is always the longest side. The theorem states a relationship between areas: the area of the square drawn on the hypotenuse equals the combined area of the squares drawn on the two shorter sides.
Proof 1: The four-triangle dissection
This proof requires only the formula for the area of a triangle (½ × base × height).
Set-up: take 4 identical right-angled triangles with legs a and b and hypotenuse c. Arrange them so their hypotenuses form the sides of an inner square, and the four outer right-angle corners of the triangles fill in an outer square.
Outer square: the four triangles are arranged inside a square of side (a + b).
Area of outer square = (a + b)² = a² + 2ab + b²
What fills the outer square? The 4 triangles plus a central square whose sides are each hypotenuse c.
Area of 4 triangles = 4 × ½ab = 2ab
Area of inner (hypotenuse) square = c²
So: a² + 2ab + b² = 2ab + c²
Subtract 2ab from both sides:
a² + b² = c² ✓
This proof is complete and rigorous — it uses only algebra and the area formula.
Proof 2: Similar triangles
This proof uses the fact that drawing the altitude from the right angle to the hypotenuse creates two smaller triangles, each similar to the original.
Set-up: in right-angled triangle ABC, the right angle is at C. Drop a perpendicular from C to the hypotenuse AB, meeting it at point D. Let AD = p and DB = q, so p + q = c.
Establishing similarity:
- Triangle ACD has angle A and a right angle at D, so its third angle = 90° − A.
- Triangle ABC also has angle A and a right angle at C. They share angle A and both have a 90°, so triangle ACD ∼ triangle ABC (AA similarity).
- Similarly, triangle CBD ∼ triangle ABC.
From triangle ACD ∼ triangle ABC:
Corresponding sides: AC/AB = AD/AC
b/c = p/b → b² = pc
From triangle CBD ∼ triangle ABC:
BC/AB = DB/BC
a/c = q/a → a² = qc
Adding:
a² + b² = qc + pc = c(p + q) = c × c = c² ✓
Proof 3: The area rearrangement (Bhaskara's proof)
Bhaskara, a 12th-century Indian mathematician, gave this elegant one-sentence proof accompanied by the word "Behold!".
Draw a square of side c (the hypotenuse). Place four copies of the right-angled triangle (legs a and b) inside the square so their hypotenuses align with the four sides of the large square. The four triangles leave a small square in the centre.
Area of the c × c square = 4 × (½ab) + area of inner square
c² = 2ab + inner square area
What is the inner square? Its sides each have length (b − a) — because each side of the inner square spans the remaining gap between the two legs of adjacent triangles placed in opposite corners.
c² = 2ab + (b − a)² = 2ab + b² − 2ab + a² = a² + b² ✓
This proof works for any positive a and b, including b > a.
Which proof is expected at GCSE?
At GCSE, the four-triangle dissection (Proof 1) is the most commonly expected proof because:
- It requires only the perfect square expansion and the area of a triangle.
- It is straightforward to write as a logical argument.
- It is a geometric proof, matching the theorem's geometric statement.
The similar-triangles proof (Proof 2) is more appropriate for GCSE Higher students who have studied similarity in depth.
How should you present a proof in a GCSE exam?
A GCSE proof should be structured as a logical argument:
- State what you are going to show ("I will show that a² + b² = c²").
- Set up the diagram with labels — define a, b, c clearly.
- State each fact and why it is true — give a reason for every step.
- Conclude with the exact statement to be proved.
| Step | Statement | Reason |
|---|---|---|
| 1 | Outer square has side (a + b) | Construction |
| 2 | Area of outer square = (a+b)² = a²+2ab+b² | Expanding perfect square |
| 3 | Area = 4 triangles + inner square = 2ab + c² | Triangle area = ½ab |
| 4 | a²+2ab+b² = 2ab+c² | Equating areas |
| 5 | a² + b² = c² | Subtracting 2ab from both sides |
Frequently asked questions
Is this the same as proving the converse of Pythagoras?
No — the converse states that if a² + b² = c² then the angle opposite c is 90°. These proofs show that a right angle implies a² + b² = c². The converse requires separate reasoning (it follows from the uniqueness of the triangle given three sides).
How many proofs of Pythagoras' theorem are there?
Hundreds — the mathematician Elisha Scott Loomis catalogued 370 distinct proofs in 1927, and more have been discovered since. The four-triangle dissection, the similar-triangles proof and the rearrangement proof are among the simplest and most widely taught.
Does Pythagoras' theorem work in three dimensions?
A 3D version gives the space diagonal of a cuboid: d² = a² + b² + c², where a, b and c are the three edge lengths. This is applied by using Pythagoras twice: first find the base diagonal, then use it as one leg of a second right-angled triangle to find the space diagonal.
Who was Pythagoras?
Pythagoras of Samos was a Greek philosopher and mathematician (c. 570–495 BCE). Although the theorem bears his name, evidence of the relationship between sides of a right-angled triangle appears in Babylonian mathematics over 1000 years before his birth. His school is credited with the first general proof, not the original discovery.
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