# Trigonometry/Proof: Heron's Formula

* Example: Triangle with sides 5-6-7
* We know that a triangle with sides 3, 4 and 5 is a right triangle. Two such triangles would make a rectangle with sides 3 and 4, so its area is ${\frac {3\cdot 4}{2}}=6$ .
* A triangle with sides 5,6,7 is going to have its largest angle smaller than a right angle, and its area will be less than ${\frac {5\cdot 6}{2}}=15$ . Let's see how much by, by calculating its area using Heron's formula.
* $s={\frac {5+6+7}{2}}=9$
* $A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {9(9-5)(9-6)(9-7)}}={\sqrt {9(4)(3)(2)}}={\sqrt {216}}=6{\sqrt {6}}\approx 14.7$
* So it's not a lot smaller than the estimate.
    $A={\sqrt {s(s-a)(s-b)(s-c)}}$
    $s={\frac {a+b+c}{2}}$

## Do you need the proof?

For most exams you do not need to know this proof. You can skip over it on a first reading of this book.
It is here for two reasons.
      * It is good practice in rather more involved algebra than you would normally do in a trigonometry course. Keep a cool head when following the steps. Most courses at this level don't prove it because they think it is too hard.
      * The proof shows that Heron's formula is not some new and special property of triangles. It has to be that way because of the Pythagorean theorem.

## Proof
### Video Link

There are videos of this proof which may be easier to follow at the Khan Academy:
* Proof of Heron's formula part I
* Proof of Heron's formula part II

### The Proof

    $A={\sqrt {s(s-a)(s-b)(s-c)}}$
    $s={\frac {a+b+c}{2}}$
The area A of the triangle is made up of the area of the two smaller right triangles.
    $A={\frac {dh+(c-d)h}{2}}={\frac {ch}{2}}$
    $A^{2}={\frac {c^{2}h^{2}}{4}}={\frac {c^{2}(b^{2}-d^{2})}{4}}={\frac {c^{2}b^{2}-c^{2}d^{2}}{4}}$
The second step is by Pythagoras Theorem.
To get closer to the result we need to get an expression for $c^{2}d^{2}$ somehow, that does not involve d or h. There is a useful trick in algebra for getting the product of two values from a difference of squares. We can get cd like this:
    $(c+d)^{2}-(c-d)^{2}=c^{2}+2cd+d^{2}-(c^{2}-2cd+d^{2})=4cd$
It's however not quite what we need. On the left we need to 'get rid' of the d, and to do that we need to get the left hand side into a form where we can use one of the Pythagorean identities for a^2 or b^2. Some experimentation gives:
    $c^{2}+2cd+d^{2}-(c-d)^{2}=4cd$ next subtract 2cd from both sides
    $c^{2}+d^{2}-(c-d)^{2}=2cd$ next use Pythagoras for a
    $c^{2}+d^{2}-(a^{2}-h^{2})=2cd$
    $c^{2}+d^{2}-a^{2}+h^{2}=2cd$ next use Pythagoras for b
    $c^{2}+b^{2}-a^{2}=2cd$
We have made good progress. We have a formula for cd that does not involve d or h. We now can put that into the formula for A so that that does not involve d or h.
    $A^{2}={\frac {4c^{2}b^{2}-(c^{2}+b^{2}-a^{2})^{2}}{16}}$
Which after expanding and simplifying becomes:
    $A^{2}={\frac {2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}-a^{4}-b^{4}-c^{4}}{16}}$
This is very encouraging because the formula is so symmetrical. We want a formula that treats a, b and c equally.
We've still some way to go. This formula is in terms of a, b and c and we need a formula in terms of s.
One way to get there is via experimenting with these formulae:
    $4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$
    $4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc$
    $4(s-a)(s-b)=-a^{2}-b^{2}+c^{2}+2ab$
Having worked those three formulae out the following complete table follows by symmetry:
    $4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$
    $4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc$
    $4s(s-b)=a^{2}-b^{2}+c^{2}+2ac$
    $4s(s-c)=a^{2}+b^{2}-c^{2}+2ab$
    $4(s-a)(s-b)=-a^{2}-b^{2}+c^{2}+2ab$
    $4(s-a)(s-c)=-a^{2}+b^{2}-c^{2}+2ac$
    $4(s-b)(s-c)=a^{2}-b^{2}-c^{2}+2bc$
Then multiplying two rows from the above table:
    $4s(s-a)\times 4(s-b)(s-c)=(-a^{2}+b^{2}+c^{2}+2bc)\times (a^{2}-b^{2}-c^{2}+2bc)$
On the right hand side of the = we have an expression that is like $(-q+p)\times (q+p)$ which is $-(q^{2})+p^{2}$ . That's a shortcut to calculating it. We could just multiply it all out, getting 16 terms and then cancel and collect them to get:
    $16s(s-a)(s-b)(s-c)=-(-a^{2}+b^{2}+c^{2})^{2}+(2bc)^{2}$
    $=-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}-2b^{2}c^{2}+4b^{2}c^{2}$
    $=-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}$
    $=16A^{2}$
and so
    $A={\sqrt {s(s-a)(s-b)(s-c)}}$
* Exercise: What about obtuse angles?
* Did you notice that just like the proof for the area of a triangle being half the base times the height, this proof for the area also divides the triangle into two right triangles? It has exactly the same problem - what if the triangle has an obtuse angle?
* Think about these three different ways we could fix the proof:
* Repeat the proof, this time with an obtuse angle and subtracting rather than adding areas.
* Choose the position of the triangle so that the largest angle is at the top. Then the problem goes away.
* Allow lengths and areas to be negative in the above proof.
* Which of those three choices is the easiest?
* Would all three approaches be valid ways to fix the proof?
* The simplest approach that works is the best. It gives you the shortest proof that is easiest to check.