## MATH 7200 : Foundations of Geometry
I

## University of Georgia, Fall
2000

## Dr. McCrory, Instructor

### A Proof of the Pythagorean Theorem
Using Heron's Formula

### by Shannon Umberger

#### Note: This proof was adapted from a proof on the Cut-the-Knot
website.

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 4A^{2} = a^{2}b^{2}. Finally,
multiplying both sides of the equation by 4 gives16A^{2}
= 4a^{2}b^{2}.

Using Heron's Formula for area, it can be stated that A^{2}
= [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, A^{2} =
[((a+b+c)/ 2)*((-a+b+c)/ 2)*((a-b+c)/ 2)*((a+b-c)/ 2)]. This implies
that 16A^{2} = (a+b+c)(-a+b+c)(a-b+c)(a+b-c). With a little
algebra, 16A^{2} = (-a^{2} + b^{2} + c^{2}
+ 2bc)(a^{2} - b^{2} - c^{2} + 2bc) =
2a^{2}b^{2} + 2a^{2}c^{2} + 2b^{2}c^{2}
- a^{4} - b^{4} - c^{4}.

Now, since 16A^{2} = 4a^{2}b^{2} and
16A^{2} = 2a^{2}b^{2} + 2a^{2}c^{2}
+ 2b^{2}c^{2} - a^{4} - b^{4}
- c^{4}, then by transitivity, 4a^{2}b^{2}
= 2a^{2}b^{2} + 2a^{2}c^{2} +
2b^{2}c^{2} - a^{4} - b^{4} -
c^{4}. Moving all terms to the left side of the equation
gives 2a^{2}b^{2} - 2a^{2}c^{2}
- 2b^{2}c^{2} + a^{4} + b^{4}
+ c^{4} = 0.

Regrouping gives (a^{4} + 2a^{2}b^{2}
+ b^{4}) - 2a^{2}c^{2} - 2b^{2}c^{2}
+ c^{4} = 0. Factoring gives (a^{2} + b^{2})^{2}
- 2c^{2}(a^{2} + b^{2}) + c^{4}
= 0. Factoring once again gives [(a^{2} + b^{2})
- c^{2}]^{2} = 0.

Therefore, a^{2} + b^{2} - c^{2} =
0, and so a^{2} + b^{2} = c^{2}, which
is the Pythagorean Theorem, which was to be proved. QED.

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