Let:
thus:
We apply Pythagoras in the triangle in order to extract :
in order to obtain one equation.
We apply Pythagoras in the triangle and obtain:
in order to obtain another equation.
Then we substitute from the first equation into the second and obtain:
So, by merging the two equations we now have a new equation in , , and and we just need to extract in order to hopefully reduce the equation to Heron's formula.
Let's collect and get rid of the square root since it's a pain:
and then let's collect for :
Now let's get to a common denominator:
We know that the permitter is expressed as so we need to shuffle the timers in some form of equation where we can express ourselves in terms of instead of , and :
Now, from the rule of polynomials we know that:
so, we apply that to the nominator, and obtain:
We observe that now we have two second-order polynomials at the denominator:
and:
So we can substitute them back into the nominator of the original equation:
and again, since the polynomial rules says that:
we can expand the nominator further to:
and arrange the terms:
Now we only know that so we have to do something about the minus (-
) signs in order to get to the addition.
and we can substitute now:
and collect :
Recall that the area of the triangle can be written as:
So we can substitute into that formula:
Now, if you recall Heron's formula, it is expressed as a semi-permitter instead of , so we need to place that inside the square root and distribute it to each of the terms:
and distribute again:
Now we know that the semi-permitter is so we can substitute it into the formula:
which is Heron's formula which we had to derive.