Famous Theorems of Mathematics/Geometry/Pythagorean Theorem

The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

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, ${\displaystyle AC^{2}+BC^{2}=AB^{2}}$, or ${\displaystyle a^{2}+b^{2}=c^{2}}$ where c is the hypotenuse.

Euclid's ProofEdit

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.