If a2+b2+c2=8R2, then prove that the triangle is a right-angled triangle.
Proof:
Given, a2+b2+c2=8R2
Using Sin law, asinA=bsinB=csinC=2R
⇒(2RsinA)2+(2RsinB)2+(2RsinC)2=8R2
⇒4R2sin2A+4R2sin2B+4R2sin2C=8R2
⇒4R2sin2A+sin2B+sin2C=8R2
⇒sin2A+sin2B+sin2C=2
⇒1-cos2A2+1-cos2B2+1-cos2C2=2 ∵cos2θ=1-2sin2θ⇒sin2θ=1-cos2θ2
⇒3−(cos2A+cos2B+cos2C)=4
⇒cos2A+cos2B+cos2C=−1
⇒−1−4cosAcosBcosC=−1(Cos2A+Cos2B+Cos2C=-1-4cosAcosBcosC,whenA+B+C=180)
⇒cosAcosBcosC=0
For the above equation to be 0, either of cosA or cosB or cosC should be equal to 0.
We know that, cos90°=0.
Thus, either of angles A,B or C is 90°(right-angled).
Hence, it is proved that the triangle is a right-angled triangle.
In In ΔABC, if 8R2=a2+b2+c2, then the triangle is