In any triangle ABC, the expression $(a + b + c) (a + b - c) (b + c - a) (c + a - b)$ is equal to

Question

In any triangle ABC, give an equivalence to the expression $$(a + b + c) (a + b - c) (b + c - a) (c + a - b)$$

Can somebody help me?

Note that $$\mathrm{area\,of\,triangle}=\sqrt{s(s-a)(s-b)(s-c)}$$ where $s=(a+b+c)/2$

Answer 1

This is nothing but the Heron's formula and $s-c={1 \over 2}(a+b+c)-c=\frac 12(a+b-c)$

So, $(a+b-c)=2(s-c),(a+b+c)=2s$,$(a+c-b)=2(s-b),(c+b-a)=2(s-a)$

So the product gives $16s(s-a)(s-b)(s-c)$ is 16 times the square of the area of the triangle.

Answer 2

Sustituting $s$ in the equation, we have $$\begin{array}{rcl} (a+b+c)(a+b-c)(a+c-b)(b+c-a)&=&(2s)(2s-2c)(2s-2b)(2s-2a)\\ &=&16\cdot s(s-a)(s-b)(s-c)\\ &=&16\cdot A^2 \end{array}$$

where $A$ is the area of the triangle, given by Heron's formula, $$A=\sqrt{s(s-a)(s-b)(s-c)}$$