♪⚝Sanjiv❀Jaiswal❁⩸

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...

Proof:
Consider the circum-circle of ΔABC with center O and radius R. Then we can have three possible figures.

Where the angle A is acute in fig i, or right angled in fig ii or obtuse in fig iii.
Let O be the center of the circum-circle of ΔABC and R be the circum-radius. Join BO and produce it to meet the circle at D.

Now in figure i,
BAC = BDC = A, BCD = 90º (angle at the semicircle).

= Sin(BDC)
= SinA
 2R

In fig iii, 
1 = Sin90º = SinA
 SinA ( Since BC = BD )
 SinA 
 2R

In...

Proof:
Let ABC be a triangle with a = BC, b = CA and c = AB, then three cases are possible. The angle C is either acute or right or obtuse angle.

Draw AD perpendicular to BC ( produce BC if necessary )
In ΔABC, SinB =  for all figures
         
or, AD = cSinB  (i)

Also in ΔACD,
1 { SinC in fig (i), Sin90 = SinC in fig (ii) and Sin(π-C) = SinC in fig (iii) }

Therefore for all the figures,
SinC = 

AD = bSinC  (ii)

From (i) and (ii) we can write,

...