The Cosine Law - Proof

Let ABC be a triangle, acute angled in fig. (i), obtuse angled in fig. (ii) and right angled in fig (iii) .
BD is drawn perpendicular to AC from point B in fig(i) and fig(ii).
To prove:  

Proof:
1. When A is acute-angled as shown in fig(i),
CD = AC - AD.......................(i)
In right-angled triangle ABD, by Pythagoras theorem, 
BD² = AB² - AD² .................(ii)
Also,

In right-angled triangle BCD, by Pythagoras theorem,
BC² = BD² + CD²
BC² = AB² - AD² + CD²                                          [ from eqn (ii) ]
BC² = AB² - AD² + (AC - AD)²                               [ from eqn (i)  ]
BC² = AB² - AD² + AC² + AD² - 2.AC.AD
BC² = AB² + AC² - 2.AC.AD                                  
BC² = AB² + AC² - 2.AB.AC cos A                       [ from eqn (iii) ]
i.e.,  

2. When A is obtuse-angled as shown in fig(ii),
CD = AC + AD..................(i)
In right-angled triangle ABD, by Pythagoras theorem, 
BD² = AB² - AD² .................(ii)
Also,

In right-angled triangle BCD, by Pythagoras theorem,
BC² = BD² + CD² 
BC² = AB² - AD² + CD²                                          [ from eqn (ii) ]                       
BC² = AB² - AD² + (AC + AD)²                              [ from eqn (i)  ]
BC² = AB² - AD² + AC² + AD² +2.AC.AD
BC² = AB² + AC² + 2.AC.AD
BC² = AB² + AC² - 2.AB.AC cos A                       [ from eqn (iii) ]
i.e., 

3. When A is right-angled as shown in fig(iii),
Here, A=90⁰ 
Taking 'cos' on both sides,
cos A = cos90⁰ = 0...........(i)
In triangle ABC, by Pythagoras theorem,
BC2 = AB2 + AC2
BC2 = AB2 + AC2 - 2.AB.AC cos A                      [ from eqn (i) ]
i.e., 

Hence for all triangles, we have 
       

i.e., 
Hence, proved   
#In a similar manner, the remaining two relations of cos B and cos C can be established