Given any right triangle, such as the one above, let the legs of the triangle have lengths a and b, and let the hypotenuse of the triangle have length c.
Using the traditional area formula, it can be stated that A = ab/ 2. This implies that 2A = ab. Squaring both sides of the equation gives 4A2 = a2b2. Finally, multiplying both sides of the equation by 4 gives16A2 = 4a2b2.
Using Heron's Formula for area, it can be stated that A2 = [s(s-a)(s-b)(s-c)]. Since s = (a + b + c)/ 2, it follows that 2s = a + b + c, and so 2s - a = b + c. Adding -a to both sides of the equation gives 2s - 2a = -a + b + c. Therefore, 2(s - a) = -a + b + c, and so s - a = (-a + b + c)/ 2. Similar algebra steps give s - b = (a - b + c)/ 2 and s - c = (a + b - c)/ 2.
By substituting back into Heron's Formula, A2 = [((a+b+c)/ 2)*((-a+b+c)/ 2)*((a-b+c)/ 2)*((a+b-c)/ 2)]. This implies that 16A2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c). With a little algebra, 16A2 = (-a2 + b2 + c2 + 2bc)(a2 - b2 - c2 + 2bc) = 2a2b2 + 2a2c2 + 2b2c2 - a4 - b4 - c4.
Now, since 16A2 = 4a2b2 and 16A2 = 2a2b2 + 2a2c2 + 2b2c2 - a4 - b4 - c4, then by transitivity, 4a2b2 = 2a2b2 + 2a2c2 + 2b2c2 - a4 - b4 - c4. Moving all terms to the left side of the equation gives 2a2b2 - 2a2c2 - 2b2c2 + a4 + b4 + c4 = 0.
Regrouping gives (a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0. Factoring gives (a2 + b2)2 - 2c2(a2 + b2) + c4 = 0. Factoring once again gives [(a2 + b2) - c2]2 = 0.
Therefore, a2 + b2 - c2 = 0, and so a2 + b2 = c2, which is the Pythagorean Theorem, which was to be proved. QED.
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