 Hello everyone, welcome to Tutor Terrific. Today we are going to prove that famous equation from antiquity, the Pythagorean Theorem. This formula is used in right triangles and it says that the sum of the squares of each leg of the triangle when added together is equal to the square of the hypotenuse. It's true for every right triangle and it comes from Pythagoras in the early days of the Greek Empire. So, to do this proof, we are going to use a very, very interesting figure, which I'm going to draw for you now. Okay, here's that figure right here. If you look at this, the first glance you think it's just a smaller square within a larger square. Well, it's a little more than that. I have actually rotated the inner square in such a way and I've made sure that square is a certain size so that when I rotate it, each of its vertices will touch the sides of the outer larger square. Now what I create when I do that are two sections of each side of the larger square, one smaller one and one larger one. Each of the smaller ones are congruent to each other and each of the larger ones are congruent to each other by doing this. Now to prove that is a separate video, of course, but we are going to assume that and now begin labeling these sections of sides. How is this useful to us? Well, it creates a bunch of right triangles. Now here's how we're going to very sneakily label the sides. The short side of each of these little triangles I'm going to call A. The larger leg of each of these triangles I'm going to call B. As you can see, I've also just finished naming the sides of the larger square. So each side of the larger square is A plus B. Now the inner square, I'm going to say that all of its sides are C as they should be because it's a square. They should all be the same. Now the backbone for this proof is how we compute the area of the larger square. There are two ways to do so. Okay, way one is going to not use the inner square at all. It is going to just use the outer square. Okay, so the outer square sides are each A plus B. So the area of the square will be equal to its side squared, which is where that term comes from, to square something. Okay, let's write out what that is after we foil. As we know from foiling, you can check my other videos as well, when we square a binomial like this, the binomial means two terms. We will have the first term squared plus the product of the two terms doubled plus the second term squared. Okay, this is our formula for the area of the entire figure. Now the other way, it's a little bit more complicated, but it is totally equivalent. So we're going to add the inner square area to the area of each of these identical right triangles. So let's write out this formula for way number two. The area equals the area of the inner square. So that's side squared, C squared, plus the area of these four right triangles. They are all identical because they all have sides A, B and C, and they are right triangles. Now the area of each of them will be one-half base times height. You can think of A as the base and B as the height. Okay, let's simplify this expression. We have C squared equals, what's four times one-half? Two. So we have C squared plus two A, B. Okay, now I said that these were two ways to find the same area, the area of the full square, the big square, so I could set these two expressions equal to each other because they are finding the area of the same large square. So that means A squared plus two A, B plus B squared is equal to C squared plus two A, B. Now we look at this equation and we notice that there's something that appears on both sides. This term two times A times B is on both sides and we can cancel it and look at what's left. None other than A squared plus B squared equals C squared. We are finished with the proof. The very elegant proof, a very nifty and intuitive, well, rather non-intuitive approach with this figure, but it sure does do the trick. Alright guys, that's the proof of the Pythagorean theorem. Thank you so much for watching and this is Falconator signing out.