Deriving Heron's Formula for the Area of a Triangle

\begin{tikzpicture} % grid \draw[help lines] (-3,-3) grid (5,3); % coordinates \coordinate[label={[black]right:$B$}] (B) at (1.5,2.5); \coordinate[label={[black]left:$C$}] (C) at (-2.5,-2); \coordinate[label={[black]right:$A$}] (A) at (3,-2); % height projection \coordinate[label={[black]south east:$D$}] (D) at (1.5,-2); \drawpoint{D}{.5mm}{black} % side labels \coordinate[label={[violet]south east:$a$}] (a) at (-0.75,0.75); \coordinate[label={[green]south east:$c$}] (c) at (2.5,0); \coordinate[label={[red]south east:$h$}] (h) at (1,0); % triangle %\draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle; \draw[green, line width=.1mm] (A) -- (B); \draw[violet, line width=.1mm] (B) -- (C); \draw[blue, line width=.1mm] (A) -- (C); % perpendicular \draw[dashed, red, line width=.1mm] (B) -- (D); % braces %\drawbrace{B}{C}{6mm}{blue}{$a$}{0}{-8mm}{mirror} %\drawbrace{A}{B}{6mm}{green}{$c$}{-6mm}{6mm}{mirror} %\drawbrace{A}{C}{6mm}{cyan}{$b$}{8mm}{4mm}{} %\drawbrace{A}{D}{2mm}{red}{$h$}{-4mm}{0mm}{mirror} \drawbrace{D}{A}{2mm}{black}{$x$}{0mm}{-4mm}{mirror} \drawbrace{C}{D}{2mm}{black}{${\color{blue}b}-x$}{0mm}{-4mm}{mirror} \end{tikzpicture}

Let:

\begin{eqnarray*} \overline{BA} &=& {\color{green}c} \\ \overline{BC} &=& {\color{violet}a} \\ \overline{CA} &=& {\color{blue}b} \\ \overline{BD} &=& {\color{red}h} \\ \overline{DA} &=& x \end{eqnarray*}

thus:

\begin{eqnarray*} \overline{CD} &=& {\color{blue}b}-x \\ \end{eqnarray*}

We apply Pythagoras in the triangle $\widehat{BDA}$ in order to extract $x$:

\begin{eqnarray*} x^{2} + {\color{red}h}^{2} &=& {\color{green}c}^{2} \\ &\Leftrightarrow& \\ x &=& \sqrt{{\color{green}c}^{2}-{\color{red}h}^{2}} \end{eqnarray*}

in order to obtain one equation.

We apply Pythagoras in the triangle $\widehat{BCD}$ and obtain:

\begin{eqnarray*} ({\color{blue}b}-x)^{2} + h^{2} &=& {\color{violet}a}^{2} \\ &\Leftrightarrow& \\ ({\color{blue}b}-x)^{2} &=& {\color{violet}a}^{2} - h^{2} \\ &\Leftrightarrow& \\ {\color{blue}b}^{2} - 2{\color{blue}b}x + x^{2} &=& {\color{violet}a}^{2} - h^{2} \end{eqnarray*}

in order to obtain another equation.

Then we substitute $x$ from the first equation into the second and obtain:

\begin{eqnarray*} {\color{blue}b}^{2} - 2{\color{blue}b}\sqrt{{\color{green}c}^{2} - {\color{red}h}^{2}} + {\color{green}c}^{2} - {\color{red}h}^{2} &=& {\color{violet}a}^{2} - h^{2} \end{eqnarray*}

So, by merging the two equations we now have a new equation in ${\color{violet}a}$, ${\color{blue}b}$, ${\color{green}c}$ and ${\color{red}h}$ and we just need to extract ${\color{red}h}$ 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:

\begin{eqnarray*} {\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2} &=& 2{\color{blue}b}\sqrt{{\color{green}c}^{2} - {\color{red}h}^{2}} \\ &\Leftrightarrow& \\ ({\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})^{2} &=& 4{\color{blue}b}^{2}({\color{green}c}^{2} - {\color{red}h}^{2}) \end{eqnarray*}

and then let's collect for ${\color{red}h}$:

\begin{eqnarray*} \color{red}h}^{2} &=& {\color{green}c}^{2} - \frac{({\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})^{2}}{4{\color{blue}b}^{2}} \end{eqnarray*}

Now let's get to a common denominator:

\begin{eqnarray*} \color{red}h}^{2} &=& \frac{4{\color{blue}b}^{2}{\color{green}c}^{2} - ({\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})^{2}}{4{\color{blue}b}^{2}} \end{eqnarray*}

We know that the permitter is expressed as $P={\color{violet}a}+{\color{blue}b}+{\color{green}c}$ so we need to shuffle the timers in some form of equation where we can express ourselves in terms of $P$ instead of ${\color{violet}a}$, ${\color{blue}b}$ and ${\color{green}c}$:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{4{\color{blue}b}^{2}{\color{green}c}^{2} - ({\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})^{2}}{4{\color{blue}b}^{2}} \\ &\Leftrightarrow& \\ {\color{red}h}^{2} &=& \frac{(2{\color{blue}b}{\color{green}c})^{2} - ({\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})^{2}}{4{\color{blue}b}^{2}} \end{eqnarray*}

Now, from the rule of polynomials we know that:

\begin{eqnarray*} P^{2} - Q^{2} &=& (P + Q )( P - Q ) \end{eqnarray*}

so, we apply that to the nominator, and obtain:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{(2{\color{blue}b}{\color{green}c} + {\color{blue}b}^{2} + {\color{green}c}^{2} - {\color{violet}a}^{2})(2{\color{blue}b}{\color{green}c} - {\color{blue}b}^{2} - {\color{green}c}^{2} + {\color{violet}a}^{2})}{4{\color{blue}b}^{2}} \end{eqnarray*}

We observe that now we have two second-order polynomials at the denominator:

\begin{eqnarray*} 2{\color{blue}b}{\color{green}c} + {\color{blue}b}^{2} + {\color{green}c}^{2} &=& \\ {\color{blue}b}^{2} + 2{\color{blue}b}{\color{green}c} + {\color{green}c}^{2} &=& \\ ({\color{blue}b} + {\color{green}c})^{2} \end{eqnarray*}

and:

\begin{eqnarray*} 2{\color{blue}b}{\color{green}c} - {\color{blue}b}^{2} - {\color{green}c}^{2} &=& \\ -({\color{blue}b}^{2} - 2{\color{blue}b}{\color{green}c} + {\color{green}c}^{2})^{2} &=& \\ -({\color{blue}b}-{\color{green}c})^{2} \end{eqnarray*}

So we can substitute them back into the nominator of the original equation:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{(({\color{blue}b} + {\color{green}c})^{2} - {\color{violet}a}^{2})({\color{violet}a}^{2}-({\color{blue}b}-{\color{green}c}^{2})^{2}}{4{\color{blue}b}^{2}} \end{eqnarray*}

and again, since the polynomial rules says that:

\begin{eqnarray*} P^{2} - Q^{2} &=& (P + Q )( P - Q ) \end{eqnarray*}

we can expand the nominator further to:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{({\color{blue}b} + {\color{green}c} + {\color{violet}a})({\color{blue}b} + {\color{green}c} - {\color{violet}a})({\color{violet}a} + {\color{blue}b} - {\color{green}c})({\color{violet}a} - {\color{blue}b} + {\color{green}c})}{4{\color{blue}b}^{2}} \end{eqnarray*}

and arrange the terms:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{({\color{violet}a} + {\color{blue}b} + {\color{green}c})(-{\color{violet}a} + {\color{blue}b} + {\color{green}c})({\color{violet}a} + {\color{blue}b} - {\color{green}c})({\color{violet}a} - {\color{blue}b} + {\color{green}c})}{4{\color{blue}b}^{2}} \end{eqnarray*}

Now we only know that $P={\color{violet}a}+{\color{blue}b}+{\color{green}c}$ so we have to do something about the minus (-) signs in order to get to the addition.

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{({\color{violet}a} + {\color{blue}b} + {\color{green}c})({\color{violet}a} + {\color{blue}b} + {\color{green}c} - 2{\color{violet}a})({\color{violet}a} + {\color{blue}b} + {\color{green}c} - 2{\color{green}c})({\color{violet}a} + {\color{blue}b} + {\color{green}c} - 2{\color{blue}b})}{4{\color{blue}b}^{2}} \end{eqnarray*}

and we can substitute $P$ now:

\begin{eqnarray*} {\color{red}h}^{2} &=& \frac{P(P-2{\color{violet}a})(P-2{\color{green}c})(P-2{\color{blue}b})}{4{\color{blue}b}^{2}} \end{eqnarray*}

and collect ${\color{red}h}$:

\begin{eqnarray*} {\color{red}h} &=& \frac{\sqrt{P(P-2{\color{violet}a})(P-2{\color{green}c})(P-2{\color{blue}b})}}{2{\color{blue}b}} \end{eqnarray*}

Recall that the area of the triangle $ABC$ can be written as:

\begin{eqnarray*} A &=& \frac{{\color{blue}b}{\color{red}h}}{2} \end{eqnarray*}

So we can substitute ${\color{red}h}$ into that formula:

\begin{eqnarray*} A &=& \frac{\sqrt{P(P-2{\color{violet}a})(P-2{\color{green}c})(P-2{\color{blue}b})}}{4} \end{eqnarray*}

Now, if you recall Heron's formula, it is expressed as a semi-permitter $p$ instead of $P$, so we need to place that $\frac{1}{4}$ inside the square root and distribute it to each of the terms:

\begin{eqnarray*} A = \sqrt{\frac{P(P-2{\color{violet}a})(P-2{\color{green}c})(P-2{\color{blue}b})}{16}} &=& \\ \sqrt{\frac{P}{2}\frac{(P-2{\color{violet}a})}{2}\frac{(P-2{\color{green}c})}{2}\frac{(P-2{\color{blue}b})}{2}} \end{eqnarray*}

and distribute again:

\begin{eqnarray*} A &=& \sqrt{\frac{P}{2}(\frac{P}{2}-\frac{2{\color{violet}a}}{2})(\frac{P}{2}-\frac{2{\color{green}c}}{2})(\frac{P}{2}-\frac{2{\color{blue}b}}{2})} \end{eqnarray*}

Now we know that the semi-permitter $p$ is $p=\frac{P}{2}$ so we can substitute it into the formula:

\begin{eqnarray*} A &=& \sqrt{p(p-{\color{violet}a})(p-{\color{green}c})(p-{\color{blue}b})} \end{eqnarray*}

which is Heron's formula which we had to derive.