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Current time:0:00Total duration:9:03

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I will now do a proof for which we credit the 12th century Indian mathematician bas Cara 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 as a bit of an angle just because I think it'll make it a little bit easier on me so let me 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 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 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 that 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 for 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 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 hypotenuse of lauryl hypotenuse is hypotony hypotenuse --is they have all length C the side opposite the right angle is all ways length C so if we can show that they all the corresponding angles are the same that we know it's congruent if you have something where all the angles are the same and you have the side that is also the corresponding side that's 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 complementary we know that because they they combine to form this angle of the square this right angle if 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 it's 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 90 minus theta and 90 degrees so they all have the same exact angle so at minimum they are similar and their hypotenuse is 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 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 this length right over here that is I'll call that lowercase B and let's assume that the shorter side 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 entire area of the entire square in terms of C well that's pretty straightforward it's a C it's a C by C it's a C by C square so the area the air here the area 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 to clip off so let me just copy and paste this copy and paste so I'm going to is our original diagram and what I will now do and actually let me clear that out 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 II 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 stick that 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 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 move that over down there and now I'm going to move 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 let me 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 and I well 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 my color in is now right over there and this triangle this triangle is now right over here that Center that's Center square and 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 is the old figure I just shifted parts of it around how can we express this in terms of 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 a plus B well that by itself is kind of interesting but what we can realize is that this length right over here 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 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 if this is length a then this is length a as well if this entire bottom is a plus B then we know that that what's leftover after subtracting the a out has to be B this whole thing is a plus B this is a then this right over here is B and so the rest of this 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 all worked out and bhaskar I gave us a very cool proof of the Pythagorean theorem