IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

rudra.panda

Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

$\cos(C) = \frac{a^2+b^2-c^2}{2ab}$

by the law of cosines. From this we get the algebraic statement:

$\sin(C) = \sqrt{1-\cos^2(C)} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}$

The altitude of the triangle on base a has length bsin(C), and it follows $A\, = \frac{1}{2} (\mbox{base}) (\mbox{altitude})\\\\ = \frac{1}{2} ab\sin(C)\\\\ = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2}\\\\ = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))}\\\\ = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)}\\\\ = \frac{1}{4}\sqrt{(c -(a -b))((c +(a -b))((a +b) -c))((a +b) +c)}\\\\ = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$

Method 2:(Using Pythagorean theorem)

Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle. The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.

In the form $4 A 2 = 4 s (s - a)(s - b)(s - c)$ , Heron's formula reduces on the left to $(c h) 2$ , or

(c b) 2 - (c d) 2

using $b 2 - d 2 = h 2$ by the Pythagorean theorem, and on the right to

(s (s - a) + (s - b)(s - c)) 2

−

((s (s - a) - (s - b)(s - c)) 2

via the principle $(p + q) 2 - (p - q) 2 = 4 p q$ . It therefore suffices to show

c b = s (s - a) + (s - b)(s - c)

, and

c d = s (s - a) - (s - b)(s - c)

.

The former follows immediately by substituting $(a + b + c) / 2$ for $s$ and simplifying. Doing this for the latter reduces $s (s - a) - (s - b)(s - c)$ only as far as $(b 2 + c 2 - a 2) / 2$ . But if we replace $b 2$ by $d 2 + h 2$ and $a 2$ by $(c - d) 2 + h 2$ , both by Pythagoras, simplification then produces $c d$ as required.

Ask & Discuss Questions with Community & Experts