The similar triangles proof draws the altitude from the right angle to the hypotenuse. That one line splits the right triangle into two smaller triangles, each similar to the original. Matching their sides gives a2 = cp and b2 = cq, where p and q are the two pieces of the hypotenuse, and adding gives a2 + b2 = c(p + q) = c2.
It is the proof most geometry textbooks use once similar triangles have been taught, and it is often called the shortest proof of the theorem. Below are the setup, the reason the triangles are similar, the algebra, a numerical check and a complete two-column write-up.
Setting up: the altitude to the hypotenuse
Start with a right triangle ABC with the right angle at C. As usual, the legs are a = BC and b = AC, and the hypotenuse is c = AB. Draw the altitude from C, the segment that meets the hypotenuse at a right angle, and call its foot D. That gives three new lengths:
- h = CD, the altitude;
- p = BD, the piece of the hypotenuse next to leg a;
- q = AD, the piece of the hypotenuse next to leg b.
The proof needs D to land between A and B, so that p + q = c. It always does. If D were at B or beyond it, triangle BCD would have a right angle at D and an angle of at least 90° at B, and no triangle has two such angles. The same argument rules out D at or beyond A.
Why the three triangles are similar
Two triangles are similar (same shape, possibly different size) when two angles of one equal two angles of the other. This is the angle-angle (AA) test, and the altitude hands it to us twice:
- Triangle ACD and triangle ABC share the angle at A, and each has a right angle (at D and at C). So △ACD ∼ △ABC.
- Triangle CBD and triangle ABC share the angle at B, and each has a right angle (at D and at C). So △CBD ∼ △ABC.
Because the angles of a triangle add up to 180°, the third angles match as well: angle DCB equals angle A, and angle ACD equals angle B. That is what the marks in the first figure show.
The order of the letters matters. In △CBD ∼ △ABC, C matches A, B matches B and D matches C. Reading the sides off in that order gives this table of corresponding sides, which is the heart of the proof:
| Triangle | Side opposite the angle equal to A | Side opposite the angle equal to B | Hypotenuse |
|---|---|---|---|
| △ABC (whole) | a | b | c |
| △CBD (piece next to a) | p | h | a |
| △ACD (piece next to b) | h | q | b |
From similar triangles to a² + b² = c²
The first proportion
Corresponding sides of similar triangles are in the same ratio. Compare the first column with the last for △CBD and △ABC:
p over a = a over c ⇒ a2 = c pThe second proportion
Do the same with the middle column for △ACD and △ABC:
q over b = b over c ⇒ b2 = c qAdding them
Add the two equations and use p + q = c:
a2 + b2= cp + cq= c(p + q)= c · c = c2Key ideaEach leg squared equals the whole hypotenuse times the piece of the hypotenuse next to that leg. The two pieces add back up to the hypotenuse, and that is the Pythagorean theorem.
The two equations have a picture. a2 = cp says the square on a has the same area as a rectangle c long and p wide, and those two rectangles together make the square on c:
Check it with the 3-4-5 triangle
Example: the pieces of the hypotenuse
Take the 3-4-5 triangle: a = 3, b = 4, c = 5. Solve each proportion for the piece of the hypotenuse:
p= a2 over c = 9 over 5 = 1.8q= b2 over c = 16 over 5 = 3.2They add up to 1.8 + 3.2 = 5 = c, as they must. The altitude comes from the area, which is 1 over 2ab using the legs and 1 over 2ch using the hypotenuse as base, so h = ab/c = 12/5 = 2.4.
The pieces are 1.8 and 3.2 and the altitude is 2.4. Triangle CBD has sides 1.8, 2.4, 3, which is 0.6 × (3, 4, 5); triangle ACD has sides 2.4, 3.2, 4, which is 0.8 × (3, 4, 5).
Try the 5-12-13 triangle yourself: you should get p = 25/13 and q = 144/13, which add up to 169/13 = 13.
The complete proof, written out
| Statement | Reason |
|---|---|
| 1. In △ABC, ∠ACB = 90°; CD ⊥ AB with D on line AB. | Given; construction |
| 2. D lies between A and B, so q + p = c. | Angles A and B are acute, and a triangle cannot have two angles of 90° or more |
| 3. △CBD ∼ △ABC | AA: ∠B is shared and ∠CDB = ∠ACB = 90° |
| 4. BD over CB = CB over AB, so a2 = cp | Corresponding sides of similar triangles are proportional |
| 5. △ACD ∼ △ABC | AA: ∠A is shared and ∠ADC = ∠ACB = 90° |
| 6. AD over AC = AC over AB, so b2 = cq | Corresponding sides of similar triangles are proportional |
| 7. a2 + b2 = c(p + q) = c2 | Add 4 and 6, then use 2 |
Nothing in these steps uses the Pythagorean theorem, so the proof is not circular. The AA test and the proportionality of corresponding sides come from the theory of parallel lines and similar figures, which is developed before the theorem.
Bonus: the altitude is a geometric mean
The two pieces are similar to each other too, since both are similar to ABC. Matching the first two columns of the table for △ACD and △CBD gives
h over p = q over h ⇒ h2 = pqFor the 3-4-5 triangle, 2.42 = 5.76 = 1.8 × 3.2. Together with a2 = cp and b2 = cq, this makes up the geometric mean theorem, which is a standard shortcut for finding altitudes and hypotenuse pieces without solving the whole triangle.
Why it is often called the shortest proof
Once similar triangles are available, the proof is two proportions and one line of algebra, with no pieces to cut or rearrange. That economy is why it is often called the shortest proof. Compare the rearrangement proof, which needs a careful argument that the inner shape is a square, or Euclid's proof, which needs five separate propositions.
The shortness is partly an illusion. A lot of work hides in the phrase "corresponding sides are proportional". Proving that rule carefully, including for lengths whose ratio is not a fraction, takes the theory of proportion in Books V and VI of Euclid's Elements. Euclid knew this proof's key fact: Book VI, Proposition 8 states that the altitude from the right angle divides the triangle into triangles similar to the whole and to each other. He still chose an area proof for Book I, where similarity was not yet available. Many historians suspect that earlier Greek proofs did use proportions in this way, though none survives.
The similarity idea also leads further. Because all three triangles have the same shape, anything that grows with the square of size behaves the same way; comparing areas instead of sides gives Einstein's proof. And because AA similarity does not depend on the theorem, it is also the starting point for non-circular trigonometric proofs. For other approaches, see all the proofs on this site.
Frequently asked questions
What are p and q in the similar triangles proof?
They are the two pieces into which the altitude from the right angle cuts the hypotenuse. Here p is the piece next to leg a and q the piece next to leg b, so p + q = c. Some books call them x and y, or c − x and x; the proof is the same.
Why are the small triangles similar to the big one?
Each small triangle shares one acute angle with the big triangle and has a right angle where the altitude meets the hypotenuse. Two equal angles are enough for similarity (the AA test), because the third angles must then match too.
Is the similar triangles proof circular?
No. It uses the AA similarity test and the fact that similar triangles have proportional sides. Neither depends on the Pythagorean theorem; both come from properties of parallel lines.
How do I find the altitude to the hypotenuse?
Use h = ab/c, which comes from writing the area two ways, or h = √pq if you know the two pieces. For legs 6 and 8 the hypotenuse is 10, so h = 48/10 = 4.8.
Does the proof work for an isosceles right triangle?
Yes. When a = b, the altitude lands at the midpoint of the hypotenuse, so p = q = c/2, and both pieces are smaller 45-45-90 triangles. The equations become a2 = c · c over 2 twice, which add up to c2.
How is this different from Einstein's proof?
Both use the same altitude and the same three similar triangles. This proof compares side lengths through proportions; Einstein's version compares areas, using the fact that the areas of similar shapes grow with the square of their size.
Sources and further reading
- Euclid, Elements, Book VI, Proposition 8 (the altitude makes similar triangles) and Proposition 31 (the theorem for similar figures), ed. D. E. Joyce, Clark University.
- Elisha S. Loomis, The Pythagorean Proposition, 2nd ed., 1940 (reissued by the National Council of Teachers of Mathematics, 1968). Catalogues many proofs built on these similar triangles.
- Pythagorean theorem, Wikipedia, section on the proof using similar triangles.
- Alexander Bogomolny, Pythagorean Theorem and its many proofs, Cut the Knot.
Keep exploring
- LearnThe Geometric Mean TheoremThe geometric mean theorem: the altitude on the hypotenuse makes three similar triangles, so h² = pq, a² = pc and b² = qc. Proof, examples and constructions.
- ProofsEuclid's Proof of the Pythagorean TheoremEuclid's windmill proof of the Pythagorean theorem, Book I Proposition 47, step by step: congruent triangles, an animated shear and the converse, I.48.
- ProofsEinstein's Proof of the Pythagorean TheoremThe similar-triangle area argument linked to Albert Einstein's boyhood proof of the Pythagorean theorem, step by step, plus the dimensional-analysis version.
- ProofsProofs of the Pythagorean TheoremProofs of the Pythagorean theorem: an animated proof, how a proof differs from checking examples, all 12 proofs on this site by type, and where to start.
- LearnFind the Height of a Triangle with the Pythagorean TheoremFind the height of an isosceles or equilateral triangle with the Pythagorean theorem: h = √(s² − (b/2)²), h = (√3/2)s, area formulas and slant heights.
- ProofsTrigonometric Proofs of the Pythagorean TheoremCan trigonometry prove the Pythagorean theorem without circular reasoning? Jason Zimba's 2009 proof and the Jackson–Johnson law-of-sines proof, explained.