Give a vector based proof of Heron's formula for computing the area of a triangle

Let $S$ denote the area of a triangle with sides of lengths $a,b,c$ . Heron’s formula states that

$S = \sqrt{s (s-a)(s-b)(s-c)}, \qquad \text{where} \qquad s = \frac{a+b+c}{2}.$

We prove this formula using vectors via the following steps. Assume the triangle has vertices $O,A$ , and $B$ with

$\lVert A \rVert = a, \qquad \lVert B \rVert = b, \qquad \lVert B - A \rVert = c.$

Using the identities
$\lVert A \times B \rVert^2 = \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2, \qquad -2A \cdot B = \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2$

prove the formula

$4S^2 = a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 = \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2).$
Simplify the formula in part (a) to obtain the formula
$S^2 = \frac{1}{16}(a+b+c)(a+b-c)(c-a+b)(c+a-b),$

and use this to deduce Heron’s formula.

Proof. We know the area of the triangle (in terms of the vectors $A$ and $B$ ) is
$S = \frac{1}{2} \lVert A \times B \rVert.$

Therefore,

$\begin{align*} 4S^2 &= \lVert A \times B \rVert^2 \\ &= \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2 \\ &= a^2 b^2 - \frac{1}{4} ( \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 ) \\ &= a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 \\ &= \frac{1}{4} (4a^2 b^2 - (c^2 - a^2 - b^2)^2) \\ &= \frac{1}{4} (2ab - (c^2 - a^2 - b^2))(2ab + (c^2 - a^2 - b^2)) \\ &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \qquad \blacksquare \end{align*}$
Proof. We simplify the formula in part (a),
$\begin{align*} && 4S^2 &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2) \\ \implies && S^2 &= \frac{1}{16} ((a+b)^2 - c^2)(c^2 - (a-b)^2) \\ \implies && S^2 &= \frac{1}{16} (a+b+c)(a+b-c)(c-a+b)(c+a-b) \\ \implies && S^2 &= \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} \cdot \frac{c-a+b}{2} \cdot \frac{c+a-b}{2} \\[9pt] \implies && S^2 &= s(s-c)(s-a)(s-b) \\ \implies && S &= \sqrt{s (s-a)(s-b)(s-c)}. \qquad \blacksquare \end{align*}$

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