I will now do a proof for which

we credit the 12th century Indian mathematician, Bhaskara. So what we’re going

to do is we’re going to start with a square. So let me see if I

can draw a square. I’m going to draw it

tilted at a bit of an angle just because I think it’ll make

it a little bit easier on me. So let me do my best

attempt at drawing something that reasonably

looks like a square. You have to bear with me if it’s

not exactly a tilted square. So that looks pretty good. And I’m assuming it’s a square. So this is a right angle. This is a right angle. That’s a right angle. That’s a right angle. I’m assuming the lengths of all

of these sides are the same. So let’s just assume that

they’re all of length, c. I’ll write that in yellow. So all of the sides of the

square are of length, c. And now I’m going to construct

four triangles inside of this square. And the way I’m going to do it

is I’m going to be dropping. So here I’m going

to go straight down, and I’m going to drop a

line straight down and draw a triangle that looks like this. So I’m going to go

straight down here. Here, I’m going to

go straight across. And so since this

is straight down and this is straight across,

we know this is a right angle. Then from this

vertex on our square, I’m going to go straight up. And since this is straight up

and this is straight across, we know that this

is a right angle. And then from this

vertex right over here, I’m going to go

straight horizontally. I’m assuming that’s

what I’m doing. And so we know that this is

going to be a right angle, and then we know this is

going to be a right angle. So we see that we’ve

constructed, from our square, we’ve constructed

four right triangles. And in between,

we have something that, at minimum, looks like a

rectangle or possibly a square. We haven’t quite

proven to ourselves yet that this is a square. Now the next thing I

want to think about is whether these

triangles are congruent. So they definitely all

have the same length of their hypotenuse. All of the hypot– I don’t know

what the plural of hypotenuse is, hypoteni, hypotenuses. They have all length, c. The side opposite the right

angle is always length, c. So if we can show that all

the corresponding angles are the same, then we

know it’s congruent. If you have something where

all the angles are the same and you have a

side that is also– the corresponding side

is also congruent, then the whole

triangles are congruent. And we can show

that if we assume that this angle is theta. Then this angle right over

here has to be 90 minus theta because together they

are complimentary. We know that because

they go combine to form this angle of the

square, this right angle. And this is 90 minus theta. We know this angle

and this angle have to add up to

90 because we only have 90 left when we subtract

the right angle from 180. So we know this has to be theta. And if that’s theta, then

that’s 90 minus theta. I think you see

where this is going. If that’s 90 minus theta,

this has to be theta. And if that’s theta, then

this is 90 minus theta. If this is 90 minus

theta, then this is theta, and then this would have

to be 90 minus theta. So we see in all four

of these triangles, the three angles are theta, 90

minus theta, and 90 degrees. So they all have the

same exact angle, so at minimum, they are

similar, and their hypotenuses are the same. So we know that all

four of these triangles are completely

congruent triangles. So with that

assumption, let’s just assume that the longer

side of these triangles, that these are of length, b. So the longer side of

these triangles I’m just going to assume. So this length right over here,

I’ll call that lowercase b. And let’s assume that the

shorter side, so this distance right over here, this distance

right over here, this distance right over here, that these

are all– this distance right over here, that these

are of length, a. So if I were to say this

height right over here, this height is of length–

that is of length, a. Now we will do

something interesting. Well, first, let’s think about

the area of the entire square. What’s the area of the

entire square in terms of c? Well, that’s pretty

straightforward. It’s a c by c square. So the area here is

equal to c squared. Now, what I’m going

to do is rearrange two of these triangles

and then come up with the area of that other

figure in terms of a’s and b’s, and hopefully it gets us

to the Pythagorean theorem. And to do that, just so we

don’t lose our starting point because our starting

point is interesting, let me just copy and

paste this entire thing. So I don’t want it to clip off. So let me just copy

and paste this. Copy and paste. So this is our original diagram. And what I will now

do– and actually, let me clear that out. Edit clear. I’m now going to shift. This is the fun part. I’m going to shift this

triangle here in the top left. I’m going to shift it below this

triangle on the bottom right. And I’m going to attempt to do

that by copying and pasting. So let’s see how much–

well, the way I drew it, it’s not that– well,

that might do the trick. I want to retain a little

bit of the– so let me copy, or let me actually cut it,

and then let me paste it. So that triangle I’m going

to stick right over there. And let me draw in the

lines that I just erased. So just to be clear, we

had a line over there, and we also had this

right over here. And this was

straight up and down, and these were

straight side to side. Now, so I moved this

part over down here. So I moved that over down there. And now I’m going to move

this top right triangle down to the bottom left. So I’m just rearranging

the exact same area. So actually let me just

capture the whole thing as best as I can. So let me cut and

then let me paste. And I’m going to move

it right over here. While I went through

that process, I kind of lost its floor,

so let me redraw the floor. So I just moved it

right over here. So this thing,

this triangle– let me color it in– is

now right over there. And this triangle is

now right over here. That center square, it is a

square, is now right over here. So hopefully you can appreciate

how we rearranged it. Now my question for

you is, how can we express the area of

this new figure, which has the exact same

area as the old figure? I just shifted

parts of it around. How can we express this in

terms of the a’s and b’s? Well, the key insight

here is to recognize the length of this bottom side. What’s the length of this

bottom side right over here? The length of this bottom

side– well this length right over here is b, this length

right over here is a. So the length of this

entire bottom is a plus b. Well that by itself is

kind of interesting. But what we can realize is that

this length right over here, which is the exact same thing

as this length over here, was also a. So we can construct

an a by a square. So this square right

over here is a by a, and so it has area, a squared. Let me do that in a color

that you can actually see. So this has area of a squared. And then what’s the area

of what’s left over? Well if this is length, a, then

this is length, a, as well. If this entire

bottom is a plus b, then we know that

what’s left over after subtracting

the a out has to b. If this whole thing

is a plus b, this is a, then this

right over here is b. And so the rest of this

newly oriented figure, this new figure, everything

that I’m shading in over here, this is just a b by b square. So the area here is b squared. So the entire area

of this figure is a squared plus b

squared, which lucky for us, is equal to the area of this

expressed in terms of c because of the exact same

figure, just rearranged. So it’s going to be

equal to c squared. And it all worked out,

and Bhaskara gave us a very cool proof of

the Pythagorean theorem.

Thanks Sal!

is that the same one vihart showed??

Your videos are so nice! It's good to watch them, we learn some interesting things in a really intuitive way… Thank you, man!

Are these video's going to be on the Khan Academy app on release?

Some men just want to proof Pythagorean Theorem.

My last name is khan. Lol.

great video.

That was awesome to watch.

mind blasted

love it!

alternately, you can find the area of the 2 rectangles (2ab) and add the area of the square in the middle, which is (b-a)^2

Works out 😉

I think so! Vi and Sal should do a collaboration!!

This proof is the easiest to understand visually, as the two areas are right next to each other and you can see what they make together.

I vaguely remember that this was given to us as a proof of the Pythagorean Theorem during high school and I never really got it. I just accepted that the Theorem is true. After watching this, I can now see how simple the explanation is. I guess it's the way you deliver it. Thanks!

stfu

this is fuckin awesome

Brilliant

Salman. I love you man, but don't you have a line tool?

something like hold down shift and it draws a vertical line, like the line tool in microsoft word would be nice

so well explained.! Thank you! I read about this and thought about checking it out.

I bet he could even just use a ruler on his notepad

excellent

I do not want to seem off topic, but you should make more videos on Biology.

I know that you specialize in mathematics, but your biology videos have helped me the most of all of them…

Jai Hind! Yeah India!!

That was so cool!

Beautiful proof

Pretty cool

Thank you!! know I know where the origin of pythagorean theorem!

can u plz tell me which software or writing pad u use for such nice video tutorials….

=D

Thank you! All other proofs I have seen have been very unsatisfactory for me!

smooth draw

Deshi, deshi bhaskara?

A law is defined; a theorem is proved.

It's probably MS Paint. :p

That's when you FEEL real education!

line tool is too mainstream now XD

Really like your vids, in the future any chance of subtitles ?

WOW., Thanks!! .. Tomorrow i have my Discrete Mathematics exam & this particular proof is coming!!!!!!!!! 😀

8:30 Sal goes Italian :0

*prove the

Mind blowing 😛

I thought the construction began by dividing each side of the larger square in the ratio a:b where a was, for argument's sake, > b and proceeded from there, with the hypotenuses forming the central square and no rearrangement required.

I'm curious, what equipment do you use for these videos?

When I refer to laws I'm really talking about axioms. An axiom is an assumption serving as a starting point for deduction. For example, if we assume that two points define a unique line, that would be an axiom. We need axioms to prove theorems.

Suppose we wanted to prove a conjecture A, and we use B to prove A. To prove B, we need C. To prove C, we need D. In order to end this chain of logic, some statement has to be taken for granted. That's an axiom. Anything we prove with axioms is a theorem.

Basically, an axiom is a statement taken for granted. A theorem is proved using axioms and other theorems.

Camtasia studio and smoothdraw3 together with a Bamboo tablet.

You can just say that each triangle has area ab/2 and there are 4 of them so 2ab in total. The central square has area (b-a)^2. So c^2 = 2ab + (b-a)^2 = a^2 + b^2. Done.

Oh, cool. Thanks.

It is much more interesting than Garfield's proof.

@jocrhru m yes i definitely mean it. my entires family is damn happy about this news. but i can show you now, my step mom used to receive an amount every month for taking some tests and surfing web. i found it here —> bit.ly/QvCJTZ?=juwxrif

any dream theater fans? anybody think this dude sounds EXACTLY like mike portnoy?

Pythagoran theorem has been pushed back to 4000+ BCE

adding 2000 years!!!

IF you do a seach of:

>>> ox theory of everything raphael <<<

scroll down to post #10

you find your evidence

stfu u fat cat

Precision isn't important, it's the idea that's important! I like this rugged style!

Brilliant.

great

This video is homophobic and it offended me

Thanks to The Great Mathematician Bhaskarya Charya and you to explain with patience

I remember this proof differently at least in the way the triangles are constructed – they divide the sides of the square in the ratio a:b and run to the opposite corner, such that the right angles of the triangles are the corners of the square and the hypotenuses are the sides of the interior square (which is proved to be a square by the law of angles in a semicircle), and no assumptions need be made. But the spirit of the proof is much the same.

Ahoy Bhashkara, ingenious.

Mathematician Bhaskara is great

Thanks for the video

Mind Blown 🙂

Baudhyana's theorem

Waste khan academy…..ntng sort

Great

This proof seems to tack on unnecessary steps. Once you have the main figure that you copied, you have a c x c square composed of 4 right triangles with legs a, b and a square with side b-a. c² = 4(1/2*a*b)+(b-a)² = 2ab + b² – 2ab + a² = a²+b².

But then you cut and move the triangles and section off the squares. It feels like you're about to make a touchdown, then you stop short, turn around, and run back 20 yards before reversing again and running for the score.

Still cool, though.

Because of western world domination,they had taken credit of all eastern world inventions. Why not we called phythogoras theoreom as bhaskara theorom

this is similar to the proof provided by Pythagoras in 300 BCE…. it's not

exactlythe same, but it's still a proof by rearrangement.0:06 "BHASKARA" not "BAASKARA"

That's what is visual intelligence