Deep-dive proof guide
Pythagorean Theorem Proof
Why does a² + b² = c²? Four complete proofs, geometric, algebraic, similar triangles, and Garfield's trapezoid, each explained step by step with diagrams.
Prerequisites
What You Need to Know First
All four proofs on this page start from the same statement: in a right triangle with legs
a and b and hypotenuse c, we want to show that:
a² + b² = c²
Each proof takes a different route to the same destination. You do not need to know trigonometry or calculus, only basic algebra and the properties of squares and triangles. If you want a refresher on the theorem itself before reading the proofs, see The Pythagorean Theorem.
Proof 1
Proof 1: The Geometric (Squares) Proof
This is the most visually intuitive proof. Draw a square on each side of the right triangle. The proof shows that the area of the large square on the hypotenuse equals the combined area of the two smaller squares on the legs.
Set up the squares
Take a right triangle with legs a and b and hypotenuse
c. Construct a square on each side: a square of area a² on leg
a, a square of area b² on leg b, and a square of area
c² on the hypotenuse c.
Observe the areas
The large square on the hypotenuse has area c². The two smaller squares have
areas a² and b² respectively.
The key claim
The claim is that the area of the large square equals the sum of the two smaller squares:
c² = a² + b²
Why this is true
This can be verified by dissection: the large square can be cut and rearranged to exactly fill the two smaller squares with no gaps and no overlaps. The interactive diagram above demonstrates this rearrangement directly. For a formal algebraic verification, see Proof 2 below.
✓ Conclusion: The area of the square on the hypotenuse equals the sum of the
areas of the squares on the two legs. Therefore c² = a² + b².
Proof 2
Proof 2: The Algebraic Rearrangement Proof
This is arguably the most elegant proof and the one most commonly shown in textbooks. It uses four copies of the right triangle arranged inside a square.
Diagram
Four triangles inside a square
The outer square has side length a + b. The four corner triangles leave a central
square whose side length is c.
Construct the outer square
Take four identical copies of the right triangle, legs a and b,
hypotenuse c. Arrange them inside a square with side length
a + b, with each triangle in one corner and hypotenuses facing inward.
Identify the inner shape
The four hypotenuses form a quadrilateral in the center. Because each hypotenuse has length
c and each corner angle is 90°, since the two acute angles of the triangle sum to
90°, this inner shape is a square with side c and area c².
Calculate the outer square's area two ways
Way 1, directly:
Outer area = (a + b)² = a² + 2ab + b²
Way 2, as inner square plus four triangles:
Outer area = c² + 4 × (ab/2) = c² + 2ab
Set the two expressions equal
a² + 2ab + b² = c² + 2ab
Simplify
Subtract 2ab from both sides:
a² + b² = c² ✓
✓ Conclusion: By calculating the area of the outer square in two different ways
and equating them, we derive a² + b² = c².
Proof 3
Proof 3: The Similar Triangles Proof
This proof uses the fact that dropping an altitude from the right angle to the hypotenuse creates two smaller triangles that are both similar to the original. It is the proof found in Euclid's Elements and is considered one of the most rigorous classical proofs.
Diagram
Altitude to the hypotenuse
The altitude from C to the hypotenuse splits the original triangle into two smaller
triangles that are similar to the original and to each other.
Draw the altitude
In right triangle ABC with the right angle at C, draw the altitude
from C perpendicular to the hypotenuse AB. Call the foot of the
altitude D.
Identify three similar triangles
This creates three triangles: triangle ABC, the original; triangle
ACD, the left portion; and triangle CBD, the right portion. All
three triangles share the same angles and are therefore similar to each other.
Write the similarity ratios
From △ABC ~ △ACD:
AC/AB = AD/AC
b/c = AD/b
AD = b²/c
From △ABC ~ △CBD:
BC/AB = DB/BC
a/c = DB/a
DB = a²/c
Use the fact that AD + DB = AB
AD + DB = c
b²/c + a²/c = c
Multiply both sides by c
b² + a² = c²
a² + b² = c² ✓
✓ Conclusion: The altitude from the right angle to the hypotenuse creates two
triangles similar to the original. Using their proportional sides and the fact that
AD + DB = c gives a² + b² = c².
Proof 4
Proof 4: Garfield's Trapezoid Proof
In 1876, James A. Garfield, then a US Congressman, later the 20th President of the United States, published an original proof of the Pythagorean theorem in the New England Journal of Education. It uses the area of a trapezoid calculated two ways.
"We think it something on which the members of both houses can unite without distinction of party."— James A. Garfield, 1876
Diagram
Garfield's trapezoid
The trapezoid contains two congruent right triangles and a central isosceles right triangle
whose two legs are both c.
Construct the trapezoid
Place two identical right triangles, legs a and b, hypotenuse
c, side by side so that one leg of each forms the two parallel sides of a
trapezoid. The parallel sides have lengths a and b, and the height
of the trapezoid is a + b.
Connect the hypotenuses
The middle triangle has two sides of length c. Because the acute angles of the
original triangle add to 90°, the angle between those two sides is a right angle. So the
middle triangle is an isosceles right triangle with legs c.
Calculate the trapezoid area directly
Area of trapezoid = (1/2) × (sum of parallel sides) × height:
Area = (1/2) × (a + b) × (a + b) = (a + b)² / 2
Calculate the trapezoid area as three triangles
The trapezoid contains two identical right triangles, each with area ab/2, and
one isosceles right triangle with legs c, area c²/2.
Area = 2 × (ab/2) + c²/2 = ab + c²/2
Set equal and simplify
(a + b)² / 2 = ab + c²/2
(a² + 2ab + b²) / 2 = ab + c²/2
a² + 2ab + b² = 2ab + c²
a² + b² = c² ✓
✓ Conclusion: By computing the area of Garfield's trapezoid in two ways, we
derive a² + b² = c². This proof is notable for being discovered by a sitting US
Congressman.
Comparison
Comparison of All Four Proofs
| Proof method | Core idea | Knowledge needed | Difficulty | Best for |
|---|---|---|---|---|
| Proof 1: Geometric (Squares) | Area of squares on each side | Basic geometry | ⭐ | Visual learners, first introduction |
| Proof 2: Algebraic Rearrangement | Four triangles inside a square | Basic algebra | ⭐⭐ | Most students, textbook standard |
| Proof 3: Similar Triangles | Altitude creates similar triangles | Triangle similarity | ⭐⭐⭐ | Geometry courses, Euclid's approach |
| Proof 4: Garfield's Trapezoid | Trapezoid area two ways | Basic algebra | ⭐⭐ | Historical interest, elegant shortcut |
There are over 370 documented proofs of the Pythagorean theorem, more than almost any other theorem in mathematics. The four above represent the main categories: visual and geometric, algebraic, similarity-based, and area-dissection. Each illuminates a different aspect of why the theorem is true.
Apply it
Apply the Theorem
Now that you understand why a² + b² = c², use the calculators below to apply it.
Find the Hypotenuse
Enter two legs and get c with full working.
Use Calculator →Find a Missing Side
Enter c and one leg to find the other.
Use Calculator →Solve Any Right Triangle
All sides, angles, area, and perimeter.
Use Calculator →FAQ
Frequently Asked Questions
Over 370 distinct proofs have been documented. Elisha Scott Loomis collected 370 of them in his 1927 book The Pythagorean Proposition. New proofs continue to be discovered; in 2023, two high school students published a proof using trigonometry that had previously been thought impossible. The theorem's simplicity and depth make it unusually fertile ground for new approaches.
Proof 2, algebraic rearrangement, is widely considered the most accessible. It requires only basic algebra and the formula for the area of a triangle. The setup, four triangles inside a square, is easy to draw and the algebra is straightforward. Proof 1 is more visual but requires an extra step to justify the dissection.
Probably, but we have no direct record. The earliest surviving proof is in Euclid's Elements, around 300 BCE, which corresponds most closely to the similar-triangles approach. Pythagoras lived around 570 to 495 BCE, and ancient sources credit him with a proof, but no original documents survive. The relationship itself was known to Babylonian mathematicians centuries before Pythagoras.
James A. Garfield, who became the 20th US President, published an original proof in 1876
while serving in Congress. It uses a trapezoid formed from two copies of the right triangle
plus a third isosceles right triangle. By computing the trapezoid's area two ways and
equating them, the proof derives a² + b² = c². It is notable for its elegance
and its unusual origin.
This was long considered circular because most trigonometric identities are derived from the Pythagorean theorem, so using them to prove it can amount to circular reasoning. However, in 2023, high school students Ne'Kiya Jackson and Calcea Johnson published a valid trigonometric proof that avoids circular reasoning, using the Law of Sines and a geometric series. Their work was published in the American Mathematical Monthly.
No, not in its standard form. In spherical geometry, the surface of a sphere, and hyperbolic
geometry, the relationship between the sides of a right triangle is different. The
Pythagorean theorem is a consequence of flat, Euclidean space. In spherical geometry, for
example, the correct formula involves cosines:
cos(c) = cos(a) × cos(b).
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