Question

How to prove heron's formula

Answer 1

The ProofEdit



Triangle used in proof.


{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}}{\displaystyle s={\frac {a+b+c}{2}}}

The area A of the triangle is made up of the area of the two smaller right triangles.

{\displaystyle A={\frac {dh+(c-d)h}{2}}={\frac {ch}{2}}}{\displaystyle 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 {\displaystyle 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:

{\displaystyle (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:

{\displaystyle c^{2}+2cd+d^{2}-(c-d)^{2}=4cd} next subtract 2cd from both sides{\displaystyle c^{2}+d^{2}-(c-d)^{2}=2cd} next use Pythagoras for a{\displaystyle c^{2}+d^{2}-(a^{2}-h^{2})=2cd}{\displaystyle c^{2}+d^{2}-a^{2}+h^{2}=2cd} next use Pythagoras for b{\displaystyle 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.

{\displaystyle A^{2}={\frac {4c^{2}b^{2}-(c^{2}+b^{2}-a^{2})^{2}}{16}}}

Which after expanding and simplifying becomes:

{\displaystyle 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:

{\displaystyle 4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}{\displaystyle 4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc}{\displaystyle 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:

{\displaystyle 4s^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}{\displaystyle 4s(s-a)=-a^{2}+b^{2}+c^{2}+2bc}{\displaystyle 4s(s-b)=a^{2}-b^{2}+c^{2}+2ac}{\displaystyle 4s(s-c)=a^{2}+b^{2}-c^{2}+2ab}{\displaystyle 4(s-a)(s-b)=-a^{2}-b^{2}+c^{2}+2ab}{\displaystyle 4(s-a)(s-c)=-a^{2}+b^{2}-c^{2}+2ac}{\displaystyle 4(s-b)(s-c)=a^{2}-b^{2}-c^{2}+2bc}

Then multiplying two rows from the above table:

{\displaystyle 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 {\displaystyle (-q+p)\times (q+p)}which is {\displaystyle -(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:

{\displaystyle 16s(s-a)(s-b)(s-c)=-(-a^{2}+b^{2}+c^{2})^{2}+(2bc)^{2}}{\displaystyle =-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}-2b^{2}c^{2}+4b^{2}c^{2}}{\displaystyle =-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}{\displaystyle =16A^{2}}

and so

{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}}

Answer 2

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