The Sine Law in terms of Circum-Radius R.Statement: In any ΔABC, where R is the Circum-Radius of ΔABC.

    a⁴  + b⁴  +c⁴  =  2c² (a²+b²)

or, (a²+b²)²  - 2a²b²  + c⁴  =  2(a²+b²)c²              [a²+ b² = (a+b)² - 2ab]

or, (a²+b²)²  - 2(a²+b²)c²   + (c2)2 =  2a²b²      

or, (a²+ b² - c2)2 =  2a²b²      

or, a²+ b² - c2=  2 ab   

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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,

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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...

a2+b2+c2=8R2sin2A+sin2B+sin2C=23(cos2A+cos2B+cos2C)=4cos2A+cos2B+cos2C=114cosAcosBcosC=1cosAcosBcosC=0cosA=0 or cosB=0 or cosC=0 A=90 or B=90 or C=90 the triangle is right angled