Proof #36
Hi! This is Pythagorean proof #36 and it is quite simple!
So, you have this figure:
As you can see, the triangle in the middle is a right triangle, and the squares are the side lengths squared (hee hee!). This is really easy to understand, and here is how it works. The Pythagorean Theorem is (a x a) + (b x b) = (c x c) (sorry I can’t do squared on this computer, though you can see what it means), so you add the two squares together. The squares added together are two green, two red, two blue and two yellow triangles. And the area of the other square is made up of two green, two red, two blue and two yellow triangles.
But first, you need to prove the triangles of the same color congruent. So the sides of the squares are parallel to each other so the line cutting the squares in half is a transversal, which by the opposite interior angles theorem makes the triangles of the same color congruent to each other in the smaller squares. The same is true for the big square. The big square's two sides are parallel, and you can make the parallel lines continue on, so by the alternate interior angles theorem, the yellow triangle in the big square is congruent to the one in the small square, and all the yellow triangles are congruent by the transitive property of congruence. You can do the exact same thing with the green triangles. Cool, huh?
There is a parallelogram with the a rectangle at its center. The sides across from each other are parallel and congruent, right? And the sides of a rectangle across from each other are congruent and parallel too, as well. So by SSS, the two red triangles are congruent and by Transitive property all the triangles are congruent. So now all the triangles are proved congruent except for the blue ones, and the blue ones in their separate squares are congruent, which makes the excess area congruent! So when you add up the two squares, all the triangles within are congruent, which makes the two smaller squares equal to the big one! This proves the Pythagorean Theorem because the squares have an area of (a x a), (b x b) and (c x c), and the Theorem is (a x a) + (b x b) = (c x c), which is what I just did, so once again, the Theorem is proved! QED
So, you have this figure:
As you can see, the triangle in the middle is a right triangle, and the squares are the side lengths squared (hee hee!). This is really easy to understand, and here is how it works. The Pythagorean Theorem is (a x a) + (b x b) = (c x c) (sorry I can’t do squared on this computer, though you can see what it means), so you add the two squares together. The squares added together are two green, two red, two blue and two yellow triangles. And the area of the other square is made up of two green, two red, two blue and two yellow triangles.
But first, you need to prove the triangles of the same color congruent. So the sides of the squares are parallel to each other so the line cutting the squares in half is a transversal, which by the opposite interior angles theorem makes the triangles of the same color congruent to each other in the smaller squares. The same is true for the big square. The big square's two sides are parallel, and you can make the parallel lines continue on, so by the alternate interior angles theorem, the yellow triangle in the big square is congruent to the one in the small square, and all the yellow triangles are congruent by the transitive property of congruence. You can do the exact same thing with the green triangles. Cool, huh?
There is a parallelogram with the a rectangle at its center. The sides across from each other are parallel and congruent, right? And the sides of a rectangle across from each other are congruent and parallel too, as well. So by SSS, the two red triangles are congruent and by Transitive property all the triangles are congruent. So now all the triangles are proved congruent except for the blue ones, and the blue ones in their separate squares are congruent, which makes the excess area congruent! So when you add up the two squares, all the triangles within are congruent, which makes the two smaller squares equal to the big one! This proves the Pythagorean Theorem because the squares have an area of (a x a), (b x b) and (c x c), and the Theorem is (a x a) + (b x b) = (c x c), which is what I just did, so once again, the Theorem is proved! QED