# Generalized Pythagorean Equation

given: triangle with an angle **C** with adjacent sides **a** and **b** and opposite side **c**.

  * bisect **b** (or **a**) at right angle with line passing thru opposite angle
    * works easier with obtuse angles such that intersection is internal to triangle - otherwise you end up hanging on the outside of the triangle which feels awkward.
    * we choose to bisect a side other than **c** so we don't bisect angle **C**
  * label this new line **x** and the divided parts of **b** as **y** and **z**
    * **y** + **z** = **b**
  * from basic Pythagorean we can state the following:
    * x<sup>2</sup> + z<sup>2</sup> = c<sup>2</sup>
    * x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup>
  * we can eliminate x<sup>2</sup> easily
    * x<sup>2</sup> = c<sup>2</sup> - z<sup>2</sup> = a<sup>2</sup> - y<sup>2</sup>
  * isolating c<sup>2</sup>
    * c<sup>2</sup> = a<sup>2</sup> - y<sup>2</sup> + z<sup>2</sup>
  * removing one of the parts of b (since y + z = b then z = b - y)
    * c<sup>2</sup> = a<sup>2</sup> - y<sup>2</sup> + (b - y)<sup>2</sup>
  * expanding (b - y)<sup>2</sup>
    * c<sup>2</sup> = a<sup>2</sup> - y<sup>2</sup> + b<sup>2</sup> - 2 * b * y + y<sup>2</sup>
  * remove the y^2s
    * c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> - 2 * b * y
  * from SOH-CAH-TOA we can state that
    * cos(C) = y/a
    * y = a * cos(C)
  * substitute for y
    * c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> - 2 * b * a * cos(C) <-- :)