# Had Heron Known Analytic Geometry…

In my previous two posts, “An Algebraic Proof of Heron’s Formula” and “An Alternative Derivation of Heron’s Formula,” I proved Heron’s formula for the area of a triangle with three given sides.  Based on Heron’s formula, we can now prove a theorem concerning the area of any triangle in a rectangle coordinate system, namely,

The area of a triangle with vertices at $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ in a rectangle coordinate system can be expressed as $|\frac{1}{2}D|\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$

where $D$ is the determinant of matrix: $\left(\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right)$

I offer the following proof: Fig. 1

By Heron’s formula, the area of triangle in Fig. 1 $A=\sqrt{s (s-a) (s-b) (s-c)}\quad\quad\quad\quad\quad(2)$

where $a, b, c$ are three sides of the triangle and, $s=(a+b+c)/2$.

Therefore, $A^2=s(s-a)(s-b)(s-c).$

where $a^2=(x_2-x_3)^2+(y_2-y_3)^2,$ $b^2=(x_1-x_3)^2+(y_1-y_3)^2,$ $c^2=(x_1-x_2)^2+(y_1-y_2)^2.$

Let $B=|\frac{1}{2} D|$,  we have $B^2 ={|\frac{1}{2} D|}^2=(\frac{1}{2}D)^2.$

Compute $A^2-B^2$ using Omega CAS Explorer (see Fig. 2) , the result shows $A^2-B^2=0.$ Fig. 2

Since $A> 0, B\geq 0$ implies $A+B >0,$ $A^2-B^2=(A-B)(A+B)=0$ implies $A-B=0$,

i.e., $A = B.$

Hence, (1) and (2) are equivalent.

I would like to learn any other alternative proof.