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^2 + z^2 = c^2
  • x^2 + y^2 = a^2
• we can eliminate x^2 easily
  • x^2 = c^2 - z^2 = a^2 - y^2
• isolating c^2
  • c^2 = a^2 - y^2 + z^2
• removing one of the parts of b (since y + z = b then z = b - y)
  • c^2 = a^2 - y^2 + (b - y)^2

    • expanding (b - y)^2
      • c^2 = a^2 - y^2 + b^2 - 2by + y^2
    • remove the y^2s
      • c^2 = a^2 + b^2 - 2by
    • from SOH-CAH-TOA we can state that
      • cos(C) = y/a so
      • y = a * cos(C)
    • substitute for y
      • c^2 = a^2 + b^2 - 2 * b * a * cos(C) ←- :)