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.
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$\begingroup$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.
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1 $\begingroup$Consider a triangle in the 3D cartesian system whose side lengths are a, b and c. Assume the three vertices of the triangle are on three coordinate axes. The area of this triangle can be represented as the sum of three area vectors where each vector corresponds to the projections of the given triangle in each of the three planes XY, YZ and ZX respectively. Let x,y be the base and altitude respectively for the XY-plane triangle, y,z for the YZ-plane triangle and x, z for the XZ-plane triangle. By Pythagoras theorem,
$$ a^2=x^2+y^2 $$$$ b^2=y^2+z^2 $$$$ c^2=z^2+x^2 $$
The area vector of the first triangle would be
$ \frac{1}{2} xy \hat{k} $
Similarly, we can form the area vectors of the other two projected triangles. The resultant vector, i.e., the area vector of the triangle in consideration would be
$ \frac{1}{2} xy \hat{k} + \frac{1}{2} yz \hat{i} + \frac{1}{2} zx \hat{j} $.
The magnitude of this vector(in effect the area of the triangle in consideration) is given by
$$ A=\frac{1}{2} \sqrt{x^2y^2+y^2z^2+z^2x^2} $$By using the three relations we derived earlier connecting x, y and z with a, b and c as well as some expansions and cancellations, we can rewrite the above equation as$$ A=\frac{1}{4} \sqrt{2c^2a^2+2c^2b^2+2a^2b^2-b^4-a^4-c^4} $$
This is what we obtain on expanding the original heron's formula in terms of s, a, b and c after substituting$$ s=\frac{(a+b+c)}{2} $$
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