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# Proof of Heron's formula (2 of 2)

## Video transcript

in the last video I claimed that this result we got for the area of a triangle that had sides of length a B and C I claimed that this is equivalent to Heron's formula and what I want to do in this video is show you that this is equivalent to Heron's formula by essentially just doing a bunch of algebraic manipulations so the first thing we want to do let's just spring this 1/2 C under the radical sign so 1/2 C that's the same thing as the square root of C squared over 4 you take the square root of that you get 1/2 C so this whole expression is equal to instead of drawing the radical I'll just write the square root of this of C squared over 4 of C squared over 4 times all of this times I'll just copy and paste it times all of that copy and paste so times all of that and of course it has to be distributed it has to be distributed so C square root of 4 times all of that and then we have to close the square root and then this thing let me just distribute the C squared over 4 this is going to be equal to the square root this is going to be hairy but I think you'll find it satisfying to see how this could turn into something as simple as Heron's formula the square root of c square root of 4 times a squared is C squared a squared over 4 minus C squared over 4 C squared over 4 I'm just distributing this times and I'm going to write it as the numerator squared over the denominator squared so times C squared plus a squared minus B squared squared over over if I squared the denominator that's 4 C squared 4 C squared and we immediately see that C squared and that C squared are going to cancel out let me close all of the parentheses just like that and then and of course this four times that 4 that's going to result in let me write it this way that's the same thing as 4 squared and I'm instead of writing 16 you'll see while I'm right why am that now this I can rewrite as this I can rewrite this is going to be equal to the square-root i'm arbitrarily switching colors of CA over 2 squared CA over 2 squared this is the same thing as that right I'm just writing it as the whole thing squared if I squared that's the C squared a squared over 2 squared over 4 minus I'm going to write this whole thing as an expression squared so that's C squared plus a squared minus B squared over 4 and we are squaring both the numerator and the denominator we squared both the numerator and the denominator now this might look a little bit interesting to you let me make the parenthesis a slightly different color you might remember from factoring from factoring polynomials that if I have something of the form x squared minus y squared that factors into X plus y times X minus y and we're going to be using this over and over again now if you call C a over 2x and you call this whole big thing Y then we have x squared minus y squared so we can factor it so this whole thing is going to be equal to the square root of of x plus y or in this case it's C a over 2 plus the Y which is C squared plus a squared minus B squared over 4 times X minus y so this is our X C a over 2 minus all of this business over here or even better let me just say plus and then let me just write the negative so plus minus C squared minus a squared plus B squared all of that over 4 so all I did here all I did here as I said this is the same thing as this Plus this this Plus this times this minus this this minus I just put a multiplied I just had plus the negative of this so minus C squared minus a squared plus B squared all I did is that right there now let's see if we could simplify this well if we could add these fractions well we can get a common denominator CA over 2 that's the same thing as 2 CA over 4 CA over 2 that's the same thing as 2 CA over 4 just multiplying the numerator in the denominator by 2 and now we can add we can add the numerators so our whole expression is now going to be equal to the square root of of this first expression will become and I'm going to write it this way I'm going to write C squared C squared plus 2 CA plus 2 C a plus a squared plus a squared minus B squared all of that over 4 that's our first expression and then our second expression is going to become well everything is going to be over 4 so I'll just write that right now everything over 4 so we could write this as B squared V squared and then we could write this as B squared minus V squared minus C squared minus 2c a minus 2c a plus a squared plus a squared just make sure thing I think they have a minus a squared here plus times the - I have them still it's a minus a squared I have a plus 2c a over here minus times a minus that's plus 2c a I have a minus C squared here I have a minus C squared here so these two things are equivalent now the next thing we need to recognize or hopefully we can recognize is that this over here that over here this might get a little bit messy that's the same thing as C plus a squared let me write this this is equal to the square root the square root open parentheses of this over here is C plus a squared minus B squared over 4 that's that first term and then the second term this over here is the same thing as C minus a squared so that whole thing will simplify to B squared minus C minus a squared all of that all of that over 4 so we're making some headway as I told you this is a hairy problem but we're seeing some neat applications of factoring polynomials and we're seeing how a fairly bizarre looking equation can be transformed into a simpler one now we can use this exact same property we have that pattern something squared minus something else squared something squared minus something else squared so we can factor it out and I'll do it in the same line so this is going to be equal to I'm going to write a little bit small just so I don't run out of space the square root this will factor into this Plus this so C plus a plus B plus B times C plus a minus B right it's this exact same pattern that I did over here this is x squared this is y squared so times C plus a minus B all of that over 4 and then we have this one this is going to be B plus C minus a so we're going to have B plus C minus a let me scroll down to the right a little bit times right B plus C minus a right that's X plus y times B minus C minus a or that's the same thing as B minus C plus a that's the same thing as B minus this is the same thing as B minus C minus a right all right and all of that all of that over for now I can rewrite this whole expression I don't want to run out of space I can rewrite this whole expression as I could write well 4 is the product of 2 times 2 4 is the product of 2 times 2 so our whole area expression has been I arguably simplified to it equals the square root and this is really the homestretch of this right here which I could just write is a + b + C / - that's that term right there times this term times that term and let me write a simplification here C plus a minus B that's the same thing as a plus B plus C minus 2b these two things are equivalent right you have an a you have a C and then B - 2 B is going to be equal to minus B right B minus 2 B that's minus B so this is going to be this next term is going to be a plus B plus C minus 2 B over 2 or instead of writing it like that let me write this over 2 minus this over 2 and then our next term right here same exact logic that's the same thing as a plus B plus C minus 2 a minus 2 a all of that over 2 right if we if we add the minus 2 a to the a we get minus a so we get B plus C minus a these are identical things so all of this over 2 or we can split the denominators just like that over 2 and then one last term and you might already recognize rule of Heron's formula popping up I was thinking not the rule of Heron Heron's formula that term right there is the exact same thing as a plus B plus C minus 2c right you take you take to see away from the C you get a minus C and then you still have the a and the B and then all of that over to write that over 2 minus that over 2 and of course we're taking the square root of all of this stuff now if we define if we define an S to be equal to a plus B plus C over 2 then this equation simplifies a good bit this right here is s that right there is s that right there is s that right there and these and these simplify good bit to - - to be over - that's the same thing as minus B minus 2a over 2 that's the same thing as minus a minus 2 C over 2 that same thing as minus C so this whole equation for our area our area now becomes our area is now equal to I'll rewrite the square root the radical the square root of s that's that right there of s times I'll do the same colors times s minus B times s minus B times this is s minus a s minus a times and where the last one s - C s minus C and we have proved Heron's formula is the exact same thing as what we proved at the end of the last video so this was pretty pretty neat and we just had to do a little bit of hairy algebra to actually prove it