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### Course: 8th grade > Unit 5

Lesson 6: Pythagorean theorem proofs# Pythagorean theorem proof using similarity

Proof of the Pythagorean Theorem using similarity. Created by Sal Khan.

## Want to join the conversation?

- Does the Pythagorean theorem work for acute or obtuse triangles?(23 votes)
- The Pythagorean Theorem is just a special case of another deeper theorem from Trigonometry called the Law of Cosines

c^2 = a^2 + b^2 -2*a*b*cos(C) where C is the angle opposite to the long side 'c'. When C = pi/2 (or 90 degrees if you insist) cos(90) = 0 and the term containing the cosine vanishes.(37 votes)

- The longest side in a right-triangle is called a hypotenuse. Does the longest side in non-right-triangles have a name?(28 votes)
- Great question! It does not have a specific name. I know it is not the hypotenuse though because that only applies to right triangles. Sorry.(19 votes)

- Is there any way for the hypotenuse in a right triangle to
*not*be the longest side?(16 votes)- No, there is not.

The lengths of the sides of a triangle go in the same order as the angles across from them :

the biggest side is across from the biggest angle

the medium side is across from the medium angle

the smallest side is across from the smallest angle

This also means that if 2 or 3 angles are the same, the sides across from them will have the same length. You see this in isosceles and equilateral triangles.

So, if we know that the longest side has to be across from the biggest angle, and a triangle has a right angle, the other angles cannot be bigger than 90, since the angles must add up to 180 total. The other 2 angles must TOTAL 90, so each must be smaller than 90. This means that their sides will have lengths shorter than the hypotenuse.(39 votes)

- can you explain why we can add together the following, and what the meaning of this addition is?

a^2 = cd

+ b^2 = ce

I understand the proof, but the addition of the two equations is confusing to me.(11 votes)- OK, so you pretend you have any two equations:

3+5=8

4+2=6

I can combine all the numbers that are on the left side into one expression, and all the numbers on the right side into one expression. So we would have:

3+5+4+2=8+6

14=14

It makes sense, right? Sal did the same thing to the equations:

a^2+b^2=cd+ce

Hope this helps! If it doesn't, let me know ;)(26 votes)

- At3:22, he uses the ~ . Does that mean anything?(10 votes)
- 5 years late, but if anyone wants to know, the ~ sign with the = under it means that two (or more shapes) are congruent to each other.(12 votes)

- Rip my grade(11 votes)
- I am a bit confused. Can someone please explain what Sal did?(8 votes)
- Sometimes its easier if you cant understand to go to the settings tab on the right hand side and watch the video again this time slower and with captions on. i highly recommend this other then just watching a different video not related to Khan Academy bc when you do that its a different type of formula and plan this is exactly how u get stuck on a problem. Before seeking any other kind of video or formula just try watching this video slower and with subtitles(6 votes)

- I understand the proof but why is it algebraically valid to add the 2 sides of the 2 equations?

a^2 = c * d

b^2 = c * e

a^2 + b^2 = c * d + c * e

These 2 equations don't seem to have any kind of relationship...(6 votes)- Lets see if this helps by doing something simpler.

If we have a=b and c=d, then the equivalent would be a+c=b+d. So if we start with a=b, we can add the same thing to both sides and it stays the same. So a+c=b+c. Then, if c=d, we can substitute in to get a+c=b+d.

We could do the same on this by saying a^2 +b^2=c*d + b^2 (adding b^2 to both sides) and end up with a^2 + b^2=c*d + c*e (substituting c*e for b^2).(9 votes)

- I still dont quite understand how he got to that(8 votes)
- This is a general question about3:16. . How do you prove that all the triangles are similar in the first place? I understand that he used the angles and color-coding but how do you do this with actual writing?(8 votes)
- there is a theorem that proves this that i learnt in my geometry class, here is a link to explain: http://jwilson.coe.uga.edu/emt668/emat6680.folders/brooks/6690stuff/righttriangle/rightday3.html(1 vote)

## Video transcript

This triangle that we have right
over here is a right triangle. And it's a right triangle
because it has a 90 degree angle, or has a
right angle in it. Now, we call the longest
side of a right triangle, we call that side,
and you could either view it as the longest side of
the right triangle or the side opposite the 90 degree angle,
it is called a hypotenuse. It's a very fancy word
for a fairly simple idea, just the longest side of a
right triangle or the side opposite the 90 degree angle. And it's just good to
know that because someone might say hypotenuse. You're like, oh, they're just
talking about this side right here, the side longest, the side
opposite the 90 degree angle. Now, what I want
to do in this video is prove a relationship, a
very famous relationship. And you might see
where this is going. A very famous relationship
between the lengths of the sides of
a right triangle. So let's say that the length of
AC, so uppercase A, uppercase C, let's call that
length lowercase a. Let's call the length of BC
lowercase b right over here. I'll use uppercases for
points, lowercases for lengths. And let's call the length of the
hypotenuse, the length of AB, let's call that c. And let's see if we can come
up with the relationship between a, b, and c. And to do that I'm
first going to construct another line or
another segment, I should say, between
c and the hypotenuse. And I'm going to
construct it so that they intersect at a right angle. And you can always do that. And we'll call this point
right over here we'll. Call this point capital D. And if you're wondering,
how can you always do that? You could imagine rotating
this entire triangle like this. This isn't a rigorous proof,
but it just kind of gives you the general idea of
how you can always construct a point like this. So if I've rotated it around. So now our hypotenuse, we're
now sitting on our hypotenuse. This is now point
B, this is point A. So we've rotated the whole
thing all the way around. This is point C. You
could imagine just dropping a rock from point C,
maybe with a string attached, and it would hit the
hypotenuse at a right angle. So that's all we did here to
establish segment CD into where we put our point D
right over there. And the reason why
did that is now we can do all sorts of
interesting relationships between similar triangles. Because we have
three triangles here. We have triangle ADC,
we have triangle DBC, and then we have the
larger original triangle. And we can hopefully
establish similarity between those triangles. And first I'll show you that ADC
is similar to the larger one. Because both of them
have a right angle. ADC has a right angle
right over here. Clearly if this
angle is 90 degrees, then this angle is going
to be 90 degrees as well. They are supplementary. They have to add up to 180. And so they both have
a right angle in them. So the smaller one
has a right angle. The larger one clearly
has a right angle. That's where we started from. And they also both
share this angle right over here, angle
DAC or BAC, however you want to refer to it. So we can actually write
down that triangle. I'm going to start with
the smaller one, ADC. And maybe I'll shade
it in right over here. So this is the triangle
we're talking about. Triangle ADC. And I went from the blue
angle to the right angle to the unlabeled angle from the
point of view of triangle ADC. This right angle isn't applying
to that right over there. It's applying to
the larger triangle. So we could say triangle
ADC is similar to triangle-- once again, you want to
start at the blue angle. A. Then we went to
the right angle. So we have to go to
the right angle again. So it's ACB. And because they're
similar, we can set up a relationship between
the ratios of their sides. For example, we know the
ratio of corresponding sides are going to do, well, in
general for a similar triangle, we know the ratio of
the corresponding sides are going to be a constant. So we could take the ratio of
the hypotenuse of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse
over the larger one, which is a AB, AC over AB is going
to be the same thing as AD as one of the legs, AD. And just to show that, I'm just
taking corresponding points on both similar triangles,
this is AD over AC. You could look at these
triangles yourself and show, look, AD, point AD, is
between the blue angle and the right angle. Sorry, side AD is between the
blue angle and the right angle. Side AC is between the blue
angle and the right angle on the larger triangle. So both of these are
from the larger triangle. These are the corresponding
sides on the smaller triangle. And if that is confusing
looking at them visually, as long as we wrote our
similarity statement correctly, you can just find the
corresponding points. AC corresponds to AB
on the larger triangle, AD on the smaller
triangle corresponds to AC on the larger triangle. And we know that AC, we can
rewrite that as lowercase a. AC is lowercase a. We don't have any
label for AD or for AB. Sorry, we do have
a label for AB. That is c right over here. We don't have a label for AD. So AD, let's just
call that lowercase d. So lowercase d applies to
that part right over there. c applies to that entire
part right over there. And then we'll call DB,
let's call that length e. That'll just make things a
little bit simpler for us. So AD we'll just call d. And so we have a over
c is equal to d over a. If we cross multiply, you have
a times a, which is a squared, is equal to c times
d, which is cd. So that's a little bit
of an interesting result. Let's see what we can do
with the other triangle right over here. So this triangle
right over here. So once again, it
has a right angle. The larger one
has a right angle. And they both share this
angle right over here. So by angle, angle
similarity, the two triangles are going to be similar. So we could say triangle
BDC, we went from pink to right to not labeled. So triangle BDC is
similar to triangle. Now we're going to look
at the larger triangle, we're going to start
at the pink angle. B. Now we're going to
go to the right angle. CA. BCA. From pink angle to right
angle to non-labeled angle, at least from the
point of view here. We labeled it before
with that blue. So now let's set up some
type of relationship here. We can say that the ratio on
the smaller triangle, BC, side BC over BA, BC over
BA, once again, we're taking the
hypotenuses of both of them. So BC over BA is going
to be equal to BD. Let me do this in another color. BD. So this is one of the legs. BD. The way I drew it
is the shorter legs. BD over BC. I'm just taking the
corresponding vertices. Over BC. And once again, we know BC is
the same thing as lowercase b. BC is lowercase b. BA is lowercase c. And then BD we defined
as lowercase e. So this is lowercase e. We can cross
multiply here and we get b times b, which, and I've
mentioned this in many videos, cross multiplying is really
the same thing as multiplying both sides by both denominators. b times b is b squared
is equal to ce. And now we can do something
kind of interesting. We can add these two statements. Let me rewrite the
statement down here. So b squared is equal to ce. So if we add the
left hand sides, we get a squared plus b squared. a squared plus b squared
is equal to cd plus ce. And then we have a c
both of these terms, so we could factor it out. So this is going to be equal
to-- we can factor out the c. It's going to be equal
to c times d plus e. c times d plus e and
close the parentheses. Now what is d plus e? d is this length,
e is this length. So d plus e is actually
going to be c as well. So this is going to be c. So you have c times c,
which is just the same thing as c squared. So now we have an
interesting relationship. We have that a squared plus b
squared is equal to c squared. Let me rewrite that. a squared. Well, let me just do
an arbitrary new color. I deleted that by accident,
so let me rewrite it. So we've just established
that a squared plus b squared is equal to c squared. And this is just an
arbitrary right triangle. This is true for any
two right triangles. We've just established that
the sum of the squares of each of the legs is equal to the
square of the hypotenuse. And this is probably
what's easily one of the most famous
theorem in mathematics, named for Pythagoras. Not clear if he's the first
person to establish this, but it's called the
Pythagorean Theorem. And it's really the basis of,
well, all not all of geometry, but a lot of the geometry
that we're going to do. And it forms the basis of a
lot of the trigonometry we're going to do. And it's a really
useful way, if you know two of the sides
of a right triangle, you can always find the third.