In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

Let *ABC* be a right-angled triangle having the angle *BAC* right.

I say that the square on *BC* equals the sum of the squares on *BA* and *AC.*

Describe the square *BDEC* on *BC,* and the squares *GB* and *HC* on *BA* and *AC.* Draw *AL* through *A* parallel to either *BD* or *CE,* and join *AD* and *FC.*

Since each of the angles *BAC* and *BAG* is right, it follows that with a straight line *BA,* and at the point *A* on it, the two straight lines *AC* and *AG* not lying on the same side make the adjacent angles equal to two right angles, therefore *CA* is in a straight line with *AG.*

For the same reason *BA* is also in a straight line with *AH.*

Since the angle *DBC* equals the angle *FBA,* for each is right, add the angle *ABC* to each, therefore the whole angle *DBA* equals the whole angle *FBC.*

Since *DB* equals *BC,* and *FB* equals *BA,* the two sides *AB* and *BD* equal the two sides *FB* and *BC* respectively, and the angle *ABD* equals the angle *FBC,* therefore the base *AD* equals the base *FC,* and the triangle *ABD* equals the triangle *FBC.*

Now the parallelogram *BL* is double the triangle *ABD,* for they have the same base *BD* and are in the same parallels *BD* and *AL.* And the square *GB* is double the triangle *FBC,* for they again have the same base *FB* and are in the same parallels *FB* and *GC.*

Therefore the parallelogram *BL* also equals the square *GB.*

Similarly, if *AE* and *BK* are joined, the parallelogram *CL* can also be proved equal to the square *HC.* Therefore the whole square *BDEC* equals the sum of the two squares *GB* and *HC.*

And the square *BDEC* is described on *BC,* and the squares *GB* and *HC* on *BA* and *AC.*

Therefore the square on *BC* equals the sum of the squares on *BA* and *AC.*

Therefore *in right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle..*

Q.E.D.

More than a millennium before Pythagoras, the Old Babylonians (ca. 1900-1600 B.C.E) used this relation to solve geometric problems involving right triangles. Moreover, the tablet known as Plimpton 322 shows that the Old Babylonians could construct all the so-called Pythagorean triples, those triples of numbers *a, b,* and *c* such that *a*^{2} + *b*^{2} = *c*^{2} which describe triangles with integral sides. (The smallest of these is 3, 4, 5.) For more on Pythagorean triples, see
X.29.Lemma 1.

The *Zhou bi* includes a very interesting diagram known as the hypotenuse diagram.
This diagram may not have been in the original text but added by its primary commentator Zhao Shuang sometime in the third century C.E. A particular case of this proposition is illustrated by this diagram, namely, the 3-4-5 right triangle.

Place four 3 by 4 rectangles around a 1 by 1 square. A 7 by 7 square results. The four diagonals of the rectangles bound a tilted square as illustrated. The area of tilted square is 49 minus 4 times 6 (the 6 is the area of one right triangle with legs 3 and 4), which is 25. Therefore the tilted square is 5 by 5, and the diagonal of the original 3 by 4 rectangles is 5.

Zhao Shuang referred to the hypotenuse figure in a general way. He described all the ways the sides, the hypotenuse, and their squares can be found in terms of each other.

The *Zhou bi* has recently been translated into English with an excellent commentary. See *Astronomy and mathematics in ancient China: the* Zhou bi suan jing, by Christopher Cullin, Cambridge University Press, 1996.

Euclid presents a proof based on proportion and similarity in the lemma for proposition X.33. Compare it, summarized here, to the proof in I.47.

Let ABC be a right-angled triangle with a right angle at A. Draw AM perpendicular to BC.
According to VI.8,
the triangles Since triangle |

Next VI.17 converts this proportion to a statement about areas, namely, the rectangle *CB* by *BM* (which is the parallelogram *BL* in the proof of I.47) equals the square on *AB.* For the same reason the rectangle *BC* by *CM* (which is the parallelogram *CL* in the proof of I.47) also equals the square on *AC.* Therefore the sum of the two rectangles *CB* by *BM* and *BC* by *CM,* which is the square on *BC,* equals the sum of the squares on *AC* and *BC.* Q.E.D.

(Actually, the final sentence is not part of the lemma, probably because Euclid moved that statement to the first Book as I.47.)

So, although Euclid’s proof in I.47 may be more complicated than some others, we can see how it well it corresponds to a simpler proof that depends on the theories of proportion and similarity.

Propositions II.12 and II.13 consider triangles other than right triangles. In II.12 the right angle is replaced by an obtuse angle, while in II.13 the right angle is replaced by an acute angle. The resulting statements are actually geometric forms of the law of cosines.

Proposition VI.31 generalizes the figures that can be placed on the sides of the right triangle to any three similar figures instead of the three squares here in I.47.