In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides. Symbolically, for a right-angled triangle *ABC* with the right angle at *C*, , or where *c* is the hypotenuse.

## Euclid's ProofEdit

This proof uses the following:

Euclid provided this proof of the Pythagorean theorem in his *Elements*, Book I, Proposition 47.

Let *ABC* be the right-angled triangle, with the right angle at *C*. Construct the squares *ABDE*, *ACFG* and *BCHJ*, and the line *CKL* perpendicular to *AB* and *ED*.

Consider the triangles *GAB* and *CAE*. Of these triangles, the sides *GA* and *CA* are equal because they are sides of the square *ACFG*. Similarly, the sides *AB* and *AE* are equal because they are sides of the square *ABDE*. Moreover, the angles *GAB* and *CAE* are equal, because they each contain the angle *CAB* plus a right angle of a square. The triangles *GAB* and *CAE* are thus congruent.

The area of the triangle *GAB* is half that of the square *ACFG*, because they share the side *GA*, and the point *B* is collinear with the square's opposite side *FC*. (This can be seen more clearly by noting that shearing *GAB* produces *GAC*, whose area is obviously half that of *ACFG*, and because shearing preserves area, so is the area of *GAB*.)

Similarly, triangle *CAE* has an area half that of the rectangle *AKLE*, because it shares the side *AE*, and *C* is collinear with *KL*.

But *GAB* and *CAE* are equal in area. Therefore, the square *ACFG* and the rectangle *AKLE* are equal in area.

By the same argument, the square *BCHJ* is equal in area to the rectangle *BKLD*. The two rectangles *AKLE* and *BKLD* make up the square *ABDE*, which is therefore equal to the area of *ACFG* plus that of *BCHJ*.