Proof of Heron's Formula for the area of a triangle

Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:

$$A = \sqrt{p(p-a)(p-b)(p-c)},$$

where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.

Could you please provide the proof of this formula?

Thank you in advance.

Solution 1:

A simple derivation exploits the cosine theorem. We have $\Delta=\frac{1}{2}ab\sin C$, hence $$ 4\Delta^2 = a^2 b^2 \sin^2 C = a^2 b^2 (1-\cos C)(1+\cos C).\tag{1}$$ On the other hand, $ 2ab\cos C = a^2+b^2-c^2$, hence $$ 2ab(1+\cos C) = (a+b)^2-c^2 = (a+b+c)(a+b-c), \tag{2}$$ $$ 2ab(1-\cos C) = c^2-(a-b)^2 = (a-b+c)(-a+b+c),\tag{3}$$ and by multiplying $(2)$ and $(3)$ and exploiting $(1)$ $$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{4} $$ which is equivalent to $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)}\tag{5} $$ as wanted.

Alternative derivation: by considering the circumcenter $O$ and its distances from the sides we have $$2\Delta = R\sum_{cyc}a\cos A,\qquad 16\Delta^2 = \sum_{cyc}a^2\cdot 2bc(\cos A)=\sum_{cyc}a^2(b^2+c^2-a^2)\tag{6}$$ through $4R\Delta=abc$ and the cosine theorem. By rearranging the RHS of $(6)$ $$ 16\Delta^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)\tag{7} $$ immediately follows.

Solution 2:

It is actually quite simple. Especially if you allow using trigonometry, which, judging by the tags, you do. If $\alpha$ is the angle between sides $a$ and $b$, then it is known that $$ \begin{align} A &= \frac{ab\sin \alpha}{2},\\ A^2 &= \frac{a^2b^2\sin^2 \alpha}{4}. \end{align} $$ Now, $\sin^2 \alpha = 1 - \cos^2 \alpha$, and you can find $\cos \alpha$ from the law of cosines: $$ c^2 = a^2 + b^2 - 2ab \cos \alpha. $$

You just find $\cos \alpha$ from this equality, plug it into the formula for $A$ above, and Heron's formula pops up as a result.

Solution 3:

Nobody can provide the proof but many can provide a proof or perhaps many proofs.

Notice that the area must be a 2nd-degree homogeneous function of $a$, $b$, and $c$, for example, if you multiply $a$, $b$, and $c$ by $9$ then you multiply the area by $9^2=81$, etc.

Next, notice that the area must be $0$ if $a+b=c$: if the distance along one side plus the distance along another side is equal to the distance along the third side, then the three corners are on a straight line, so the area is $0$. For that reason $a+b-c$ should appear as a factor, i.e. as something you multiply by.

For the same reason, $b+c-a$ and $c+a-b$ should be factors.

Next, notice that if $a=b=c=0$, then the area must be $0$, so that's why $a+b+c$ should be a factor.

Now we have $(a+b+c)(a+b-c)(b+c-a)(c+a-b)$. That is homogeneous of degree $4$ rather than $2$, so take its square root and we have $\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$, which is homogenous of degree $2$. That, then, should be proportional to the area.

Now let's see what the constant of proportionality is: The area of a right triangle with legs of length $1$ and hypotenuse of length $\sqrt2$ is $1/2$. Plugging in those three numbers for $a$, $b$, and $c$ we get $$ \frac 1 2 =\text{constant}\times\sqrt{\left(1+1+\sqrt 2\right) \left(1+1-\sqrt 2\right)\left(1-1+\sqrt 2\right)\left(-1+1+\sqrt2\right)} = \text{constant}\times 2. $$ So the "constant" is $\dfrac 1 4$ and finally we have $$ \text{area} = \frac 1 4 \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$ That's Heron's formula.

Solution 4:

I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any, and seems to be simpler than the usual methods that are applied.

A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the sum of two side lengths , $c = m + n$, using Pythagorean theorem and the area formula of a triangle, we can prove Herons formula.

First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:

$a^2 - m^2 = h^2$

$b^2 - n^2 = h^2$

$a^2 - m^2 = b^2 - n^2$

$a^2 - b^2 = m^2 - n^2$

Then using $c = m + n$, eliminate $m$:

$a^2 - b^2 = (c-n)^2 - n^2$

$a^2 - b^2 = c^2 - 2cn$

$n = \frac{1}{2c}(c^2 + b^2 - a^2)$

We can eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:

$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$

Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:

$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $

$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $

$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $

$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $

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

Q.E.D.