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:
  • x² + z² = c²
  • x² + y² = a²
• we can eliminate x² easily
  • x² = c² - z² = a² - y²
• isolating c²
  • c² = a² - y² + z²
• removing one of the parts of b (since y + z = b then z = b - y)
  • c² = a² - y² + (b - y)²
• expanding (b - y)²
  • c² = a² - y² + b² - 2 * b * y + y²
• remove the y^2s
  • c² = a² + b² - 2 * b * y
• from SOH-CAH-TOA we can state that
  • cos(C) = y/a
  • y = a * cos(C)
• substitute for y
  • c² = a² + b² - 2 * b * a * cos(C) ←- :)