Heron's Formula

[quote]I really think this is cool.[/quote]
[b]Construction:[/b]
ΔABC with BC = a; AC = b, AB = c is constructed and AD is drawn from A such that it intersects BC at D in right angles. Now, let us take BD = d & DC = a - d and AD = h

[b]Proof:[/b]
In ΔABD, d[sup]2[/sup] + h[sup]2[/sup] = c[sup]2[/sup]
⇒ h[sup]2[/sup] = (c + d) (c - d) = c[sup]2[/sup] - d[sup]2[/sup]                                          .... (i)
In ΔACD, (a-d)[sup]2[/sup] + h[sup]2[/sup] = b[sup]2[/sup]
⇒ (a² - 2ad + d²) + (c² - d²) = b² [from (i)]
[math]\implies d = \frac{a^2 + c^2- b^2}{2a}[/math]                                                                                            ....(ii)

Now, we know that the Area of Triangle(T) = [math]\frac {ah}{2}[/math]
[math]\implies T^2 = \frac {a^2 h^2}{4}[/math]

From (i) & (ii),

[math]T^2 = [c + \frac{a^2 + c^2- b^2}{2a}][c - \frac{a^2 + c^2- b^2}{2a}] \times a^2 \times \frac{1}{4}[/math]
[math]\implies T^2 = [\frac{2ac + a^2 + c^2 - b^2}{2a}][\frac{2ac - a^2 - c^2 + b^2}{2a}] \times a^2 \times  \frac{1}{4}[/math]
[math]\implies T^2 = \frac{[(a+c)^2 - b^2][- (a - c)^2 + b^2]}{16}[/math]
[math]\implies T^2 = \frac{(a + c + b) (a + c - b) ( a + b -c) (b + c - a)}{16}[/math]
[math]\implies T^2 = \frac{2(\frac{a + b+c}{2})[2(\frac{a + b+c}{2})  - 2a][2(\frac{a + b+c}{2})  - 2b][[2(\frac{a + b+c}{2})  - 2c]}{16}[/math]

Substituting  ½(a+ b+ c) with s,
[math]T^2 = \frac{(2s)(2s-2a)(2s-2b)(2s-2c)}{16}[/math]
[math]\implies T^2 = s (s-a) (s-b) (s-c) [/math]
[math]\implies T = \sqrt{s (s-a) (s-b) (s-c)}\;\blacksquare[/math]