In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In  a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled

$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}.......(1)$
By Heron's formula,$\Delta=\sqrt{\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}}$
$\Delta^2=\frac{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}{16}$
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=16\Delta^2$
Putting in $(1)$
$16\Delta^2=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$
By formula $R=\frac{abc}{4\Delta}$,we get
$a^2+b^2+c^2=8R^2$
By sine rule,$a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\sin^2A+\sin^2B+\sin^2C=2$
I am stuck here.

$\sin^2A+\sin^2B+\sin^2C =1-\cos^2A+ \dfrac{1-\cos(2B)}{2} + \dfrac{1-\cos(2C)}{2}=1-\cos^2A+ 1 - \cos(B+C)\cos(B-C)=2-\cos^2A+\cos A\cos(B-C)=2$. From this you can see $\cos A = 0$,or $A = B-C \Rightarrow B = \dfrac{\pi}{2}$. Thus $B)$ is the answer.
