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Pythagorean theorem proof using similarity

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

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  • orange juice squid orange style avatar for user N Peterson
    Does the Pythagorean theorem work for acute or obtuse triangles?
    (21 votes)
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    • blobby green style avatar for user JohnWmAustin
      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.
      (23 votes)
  • piceratops ultimate style avatar for user Philothea
    Is there any way for the hypotenuse in a right triangle to not be the longest side?
    (11 votes)
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    • starky tree style avatar for user Sandy Knight
      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.
      (28 votes)
  • blobby green style avatar for user jamesclarks
    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.
    (6 votes)
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    • aqualine ultimate style avatar for user hannahmorrell
      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 ;)
      (13 votes)
  • hopper jumping style avatar for user :)
    At , he uses the ~ . Does that mean anything?
    (7 votes)
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  • duskpin ultimate style avatar for user Elziule
    I am a bit confused. Can someone please explain what Sal did?
    (6 votes)
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    • leaf grey style avatar for user Hannah E
      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
      (3 votes)
  • leaf orange style avatar for user Bogdan Dancu
    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...
    (4 votes)
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    • mr pink green style avatar for user David Severin
      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).
      (4 votes)
  • aqualine ultimate style avatar for user Misabelle
    Can anyone simplify what he just said? Please? I'm confused....
    (2 votes)
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  • leaf grey style avatar for user Ajan Prabakar
    This is a general question about . . 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?
    (5 votes)
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  • blobby blue style avatar for user Tetiana Kondakova
    Why would we add the two ratios ( on the video)?
    (2 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      If you have two equations that discuss the same variables, you can add them together. This is the same thing as adding something to both sides of an equation, but you just substitute one of the 'somethings' with another something that its equal to. This means the sum equation will still be equal.
      If we have 2x = 4 and 7y = 2x, we can add them together. Either side of the equation can match up with either side of the other equation, so when we add them we could get 2x + 7y = 2x + 4 or 2x + 2x = 7y + 4 (both versions are equal).
      (5 votes)
  • leafers ultimate style avatar for user Rok Wisdawn
    Nothing I remember or watched, so far, prepared me for the whole “triangle A is similar to triangle B” postulate; or with specific reference to this video, the argument of “triangle ADC is similar to triangle ACB” at . I understand that they share one side and one angle, but I still don't understand why that qualifies the “similarity” relationship. Moreover, what does it even mean when we say “triangle X is similar to triangle Y”? And when would two triangles qualify for that relationship of similarity?
    (2 votes)
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    • piceratops tree style avatar for user blobbyfishy
      Good question. If a triangle is similar to another triangle, it must have at least one same side, and at least two similar angles. In the video, when Sal says that two triangles are similar, he shows two equal angles, as well as one common side.
      I hope this helped u!
      (3 votes)

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.