# 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?

• – lab bhattacharjee Nov 22 '13 at 5:02
• Both the links go about in a bit complicated fashion. The proof is simple and with very less amount of algebraic manipulation as shown in paramanands.blogspot.com/2011/05/… Search for "Heron" in the blog post. – Paramanand Singh Nov 22 '13 at 5:13
• Relevant question asking for visual proofs here. – user26486 Jun 24 '15 at 21:50
• Calculate the perpendicular height & use brute force. – Donald Splutterwit Sep 15 '17 at 15:23
• Wikipedia shows several proofs. – John Wayland Bales Sep 15 '17 at 15:37

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.

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.

• To avoid the calculations, see WolframAlpha. – user26486 Jun 24 '15 at 21:49

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.