# Help reducing the following expression involving Heron's formula for the area of a triangle.

Let $(ABC) = \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c).}$ This is Heron's formula for $(ABC),$ the area of $\Delta ABC,$ with sides $a, b, c,$ using $s = \frac{a + b + c}{2},$ the semiperimeter.

I have the following step that I don't understand in a problem I am working using the above notation.

$$(ABC) \cdot (\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}) =$$ $$\frac{(ABC) \cdot abc}{s \cdot (s - a) \cdot (s - b) \cdot (s - c)}$$

I realize it's probably "just" algebra and suspect that the reduction uses the fact that $s - b = \frac{a - b + c}{2},$ and similarly for $s - a$ and $s - c.$

But all my attempts to show the reduction (as given in a solution) have gotten complicated and I haven't been able to derive the result.

For context, the problem is 3.2.4 in Coxeter and Greitzer's book Geometry Revisited:

In the notation of Section 1.4, $$r_a + r_b + r_c - r = 4R,$$ where $r_a, r_b, r_c$ are the excircle radii of $\Delta ABC,$ and $r, R$ are the incircle and circumcircle radii of $\Delta ABC,$ respectively.

Thanks to any that can help with this.

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$$\frac{1}{s-a}+\frac{1}{s-b}+\frac{1}{s-c}-\frac{1}{s}=\frac{s(s-b)(s-c)+s(s-a)(s-c)+s(s-a)(s-b)-(s-a)(s-b)(s-c)}{s(s-a)(s-b)(s-c)}=\frac{3s^3-(2a+2b+2c)s^2+(ab+ac+bc)s-s^3+(a+b+c)s^2-(ab+ac+bc)s+abc}{s(s-a)(s-b)(s-c)}=\frac{2s^3-(a+b+c)s^2+abc}{s(s-a)(s-b)(s-c)}=\frac{abc}{s(s-a)(s-b)(s-c)}$$
Notice for the last step we have used $(a+b+c)=2s$.