# Proof of Heron's formula (2 ofÂ 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 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
manipulation. 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
times all of this. I'll just copy and paste it. Copy and paste. So times all of that. And of course, it has
to be distributed. So c squared over 4
times all of that. And then we have to
close the square root. 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 squared
over 4 times a squared is c squared a squared over 4,
minus c squared over 4. I'm just distributing this. 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-- if I square the
denominator that's 4c 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, of course, this 4 times
that 4, that's going to result in-- well let
me write it this way. That's the same
thing as 4 squared. And I'm instead of writing
16, you'll see why I'm writing that. Now this I can rewrite. This is going to be equal to
the square root-- I'm arbitrarily switching colors--
of ca over 2 squared. This is the same thing as that. Right? I'm just writing it as
the whole thing squared. If I square that, that's the c
squared a squared over 2 squared over 4, minus-- and 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. Now this might look a little
bit interesting to you. Let me make the parentheses in
a slightly different color. You might remember 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 ca 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 x plus y, or in this case it's
ca 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. ca 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 is I said
this is the same thing as this plus this, this plus this,
times this minus this, this minus-- I just said 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 can
simplify this, or if we could add these fractions. Well, we can get a
common denominator. ca over 2, that's the same
thing as 2ca over 4. ca over 2, that's the same
thing as 2ca over 4, just multiplying the numerator
and the denominator by 2. And now we can add
the numerators. So our whole expression is now
going to be equal to the square root of this first expression,
will become-- and I'm going to write it this way. I'm going to write c squared
plus 2ca 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's going to be over
4, so I'll just write that right now. Everything over 4. And then we could write this
as b squared, minus c squared minus 2ca plus a squared. Just to make sure, I have
a minus a squared here. Plus times a minus, still
it's a minus a squared. I have a plus 2ca over here. Minus times a minus,
that's a plus 2ca. 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-- 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, 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 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. 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 times
c plus a minus b. Right? It's the 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. Let me scroll down to
the right a little bit. Times b plus c minus a--
that's x plus y-- times b minus c minus a. Or that's the same thing
as b minus c plus a. This is the same thing
as b minus c minus a. Right? All right. And all of that over 4. Now, I can rewrite this
whole expression. I don't want to
run out of space. I can rewrite this whole
expression as, well 4 is the product of 2 times 2. So our whole area expression
has been, arguably, simplified to it equals the square root--
and this is really the home stretch-- of this right here,
which I can just write as a plus b plus c over 2. 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 minus 2b is going to be equal to minus b. Right? b minus 2b,
that's minus b. So this next term is going
to be a plus b plus c minus 2b, 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 2a, all of that over 2. Right? If we add the minus 2a to
the a we get minus a. So we get b plus c minus a. These are identical things. So all this over 2, or we
can split the denominators just like that over 2. And then one last term. And you might already
recognize the 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 2c 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 2. You could 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 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. And that right there is S. And these simplify
a good bit too. Minus 2b over 2, that's just
the same thing as minus b. Minus 2a over 2, that's the
same thing as minus a. Minus 2c over 2, that's the
same thing as minus c. So this whole equation for our
area now is equal to-- I'll rewrite the square root. The radical, the square root,
of S-- that's that right there. I'll do it in the same colors. Times S minus b, times this is
S minus a, times-- and we're at the last one-- 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 neat. And we just had to do a
little bit of hairy algebra to actually prove it.