Prove: Pythagorean theorem, that is, the sum of squares of two right angles of a right triangle is equal to the square of the hypotenuse.

Proof: it is discussed in two cases, that is, the lengths of two right-angled sides are unequal and equal.

The lengths of the two right-angled sides are not equal.

As shown in the figure, let the side lengths of a right triangle be a, b, c, (a

Put four triangles of the same size together in the form of the right picture, and then:

Then the area of the big square on the right is the sum of the areas of the four right triangles and the area of the small square in the middle.

de:C2 = 4 *(ab/2)+(b-a)2 = 2ab+a2+B2-2ab = a2+B2。

That is, A 2+B 2 = C 2, the original proposition is proved.

2.？ These two right angles are equal in length.

As shown in the figure, let the right side and hypotenuse of a right triangle be a and c respectively.

Put four triangles of the same size together in the form of the right picture, and then:

Then the area of the square on the right is the sum of the areas of the four right triangles.

De: c 2 = 4 * (aa/2) = 2a 2 = a 2+a 2。

That is, a 2+a 2 = c 2, the original proposition is proved.

Therefore, the sum of squares of two right-angled sides of a right-angled triangle is equal to the square of the hypotenuse.