Proof:
↪ We consider a triangle ABC then we can have three possible figures where angle C is acute in figure i, right angle in figure ii and obtuse in figure iii.

From A draw AD perpendicular to BC ( Produce BC if necessary )

In ΔABC, AB² = AD² + BD²
⇒c² = AD² + BD²  → Ⅰ

In figure (i), 
⇒⇒ AD = bSinC

Also, 
⇒ DC = ACCosC = bCosC
⇒ BD = BC -  DC = a-bCosC

Now from equation Ⅰ,
c² = AD² + BD² (becomes)
c² = (bSinC)² + (a-bCosC)²
c² = b²Sin²C + a² - 2abCosC + b²CosC²
c² = b² + a² - 2abCosC
⇒ 2abCosC = b² + a² - c²
⇒

Similarly in figure ii,

AB² = BC² + AC²
c² = a² + b² = a² + b² - 2abCosC
2abCosC = a² + b² - c²
⇒

And in figure iii,


⇒
⇒ AD = bSinC

Also, 
⇒
⇒ CD = -bCosC
⇒ BD = BC + CD = a - bCosC

Now from equation Ⅰ,
c² = AD² + BD² 
   = (bSinC)² + (a-bCosC)²
   = b²Sin²C + a² + b²Cos²C - 2abCosC
   = b² + a² - 2abCosC
⇒ 2abCosC = b² + a² - c²
⇒

Finally in all figures, it is seen 
Similarly we can show