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.
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.