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= BC^2...

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 fig iii,SinA
SinA
2R

Therefore in each...

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 inΔACD,
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,

AD = bSinC = cSinB

Similarly we can show,


Combining these we get,
Proved.