The Sine Law

Let ABC be a triangle, acute angled in fig(i), obtuse angled in fig(ii) and right angled in fig(iii).
A perpendicular AD is drawn on BC [BC is produced to D in fig. (ii) and D coincides C in fig (iii)].
To Prove: 

Proof:
1. When C is acute-angled as shown in fig(i),
From triangle ADB,
    
From triangle ADC,
    

From eqⁿ:(i) &  eqⁿ:(ii)
    

2. When C is obtuse-angled as shown in fig(ii),
From triangle ADB,
    
From triangle ADC,
    

From eqⁿ:(i) &  eqⁿ:(ii)
    

3. When C is right-angled as shown in fig(iii),
    
Also, 
    

From eqⁿ:(i) &  eqⁿ:(ii)
    

Hence, for all cases, we have,
    ..........(I)

Similarly, when a perpendicular is drawn from B on CA, we have
    ..........(II)
Combining eqⁿ:(I) &  eqⁿ:(II) 
For all cases, we have,
                                                  , which is the Sine Law.