Формула Герона

Oh, so many formulas! And yet another formula for the area of a triangle! First, let's review.

Alright, good. Let's try something.

We know that the area of a triangle ABC is given by $\frac{1}{2}ab\sin{C}$. We don't really know a whole lot of formulas relating sides and sines of angles, but the Law of Cosines relates the sides and the cosine of an angle!

By the Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos{C}$. Rewrite this as $\cos{C} = \frac{a^2 + b^2 - c^2}{2ab}$ to get $\cos{C}$ in terms of the side lengths.

Now, let's return to our formula. Since $\sin{C} = \sqrt{1 - \cos^2{C}}$, (Click to reveal the grinding through the algebra)

$$\begin{align*} \frac{1}{2}ab\sin{C} &= \frac{1}{2}ab\sqrt{1 - \cos^2{C}} \\ &= \frac{1}{2}ab\sqrt{1 - \left(\frac{a^2 + b^2 - c^2}{2ab}\right)^2} \\ &= \frac{1}{2}ab\sqrt{1 - \left(\frac{(a^2 + b^2 - c^2)^2}{4a^2b^2}\right)} \\ &= \sqrt{\frac{4a^2b^2 - (a^2 + b^2 - c^2)^2}{16}} \\ &= \sqrt{\frac{4a^2b^2 - (c^2 - a^2 - b^2)^2}{16}} \\ &= \sqrt{\frac{(2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2)}{16}} \\ &= \sqrt{\frac{((a^2 + 2ab + b^2) - c^2)(c^2 - (a^2 - 2ab + b^2))}{16}} \\ &= \sqrt{\frac{((a+b)^2 - c^2)(c^2 - (a-b)^2)}{16}} \\ &= \sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}} \\ &= \sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{a-b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)} \end{align*}$$

Now to simplify this, we let the semiperimeter $s = \frac{a+b+c}{2}$. $$A = \sqrt{s(s-a)(s-b)(s-c)}$$

Phew. This is our new formula, known as Heron's formula!! It can be particularly useful if you are given all the sides of a triangle.

There are many formulas for the area of a triangle, and one, Heron's Formula, relates the area of a triangle to the lengths of the three sides.

The formula is: for a triangle $\triangle ABC$, where the length of $AB$ is $c$, the length of $AC$ is $b$, and the length of $BC$ is $a$, $[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$.

This will only be one of the many proofs of Heron's Theorem. Let's first start experimenting--we could draw the incircle, as it might help us.

So looking back at our figure, we notice some equal tangents. $AE \cong AF$, $BF \cong BD$, and $CD \cong CE$. If we denote the lengths of these segments $x$, $y$, and $z$, respectively, then we can see $2x+2y+2z=p$, where $p$ is the perimeter of $\triangle ABC$. We arrive at $x+y+z=s$.

We can see that $c=x+y$, and $BD=y$, so using the fact that $x+y+z=s$, we have $y=x+y+z-(x+z)=s-b$ so we have expressed the length in terms of $a$, $b$, and $c$. Using the same method for the lengths $x$ and $z$, we have $x=s-a$ and $z=s-c$, as desired.

We can note now that $[ABC]=[AIE]+[AIF]+[BID]$ $+[BIF]+[CIE]+[CID]=\frac{1}{2}r(s-a)+$ $\frac{1}{2}r(s-a)$ $+\frac{1}{2}r(s-b)+\frac{1}{2}r(s-b)+\frac{1}{2}r(s-c)+\frac{1}{2}r(s-c)=$ $r(s-a)+r(s-b)+r(s-c)=$$r(s-a+s-b+s-c)=r(3s-2s)=rs$. $\blacksquare$

One thing that can really help us right now is: the excircles of $\triangle ABC$.

How, you ask? Well, let's see.

Here we can note something important. We can use ratios to complete the rest of the proof.

We can notice that $\triangle IAF$ and $\triangle XAP$ are similar because of $AA$. Thus, noting that the sides of similar triangles are in ratio, $\frac{r}{s_a}=\frac{r_a}{s}$, and rewriting gives $$rs=r_a(s-a)$$ Because we know $rs$ is the area of a triangle, we have alternately discovered another formula for the area of a triangle: $$[ABC] = r_a(s-a)$$ However, this is not necessary for this proof.

We will now prove that $XB$ and $XC$ are the external angle bisectors of $\triangle ABC$. $\triangle XUC \cong \triangle XPC$ because of $SSS$, so $\angle UCX \cong \angle PCX$. We can use the same argument for $XB$, so we can conclude that the lines connecting the base angles to the excenter bisect the external angles.

Having that piece of information, we set $m\angle ABF = m\angle HBC = x$. $m\angle XBU = \frac{180-2x}{2} = 90-x$, so $m\angle BXU = x$. We now see that $\triangle IBU \sim \triangle XBU$.

Now we can notice that $$AB+BQ=AC+CP$$ because of equal tangents. Because $BQ=BU$ and $CP=CU$, we can substitute to obtain $$AB+BU=AC+CU$$ Since $AB+BU+AC+CU=AB+BC+AC=2s$, we have $AB+BU=AC+CU=s$. We can now see that $BU=s-c$.

We have already seen that $m\angle BXU = m\angle IBU$, so using $AA$ similarity we see that $\triangle IBD \sim \triangle XBU$. Using the fact that corresponding sides of similar triangles are in ratio, we have $\frac{r}{s-b}=\frac{s-c}{r_a}$. Rearranging gives $$rr_a=(s-b)(s-c)$$

Now we are very close to finishing.

We have the two relations: $$rs=r_a(s-a)$$ and $$rr_a=(s-b)(s-c)$$ Multiplying these two together gives $$r^2sr_a=r_a(s-a)(s-b)(s-c)$$ Simplifying gives $$r^2s=(s-a)(s-b)(s-c)$$, and since we want to express the right hand side as the area of the triangle, we multiply both sides by $s$ and take the square root to finally get $$rs=[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$$ As desired. $\blacksquare$