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.
 A: $\sin^2A+\sin^2B+\sin^2C =1-\cos^2A+ \dfrac{1-\cos(2B)}{2} + \dfrac{1-\cos(2C)}{2}=1-\cos^2A+ 1 - \cos(B+C)\cos(B-C)=2-\cos^2A+\cos A\cos(B-C)=2$. From this you can see $\cos A = 0$,or $A = B-C \Rightarrow B = \dfrac{\pi}{2}$. Thus $B)$ is the answer.
A: You have
$\sin^2A+\sin^2B+\sin^2C=2
$.
What stands out to me
is that this is true
if the triangle is right:
$\sin C = 1$
and
$\sin^2A+\sin^2B
=\sin^2A+\sin^2(\pi/2-A)
=\sin^2A+\cos^2(A)
=1
$.
A: Without loss of generality suppose $\cos C\ne 0 $.
Let the area of the triangle be T. The LHS is $16 T^2. $
The RHS is $8(\frac {1}{2}(a b \sin C)^2) (4\sin^2 C)c^2/(a^2+b^2+c^2)=32 T^2c^2/(a^2+b^2+c^2).$
And  the denominator above is $a^2+b^2+c^2=(c^2+2 a b \cos C)+c^2=2(c^2+a b \cos C).$ So we have $$1=c^2/(\sin^2 c)(c^2+a b \cos C)\iff$$  $$\iff c^2=c^2\sin^2 C+a b \cos C \sin^2 C$$  $$\iff c^2\cos^2 C=a b \cos C\sin^2 C\iff$$ $$ \iff c^2\cos C=a b \sin^2 C=a b (1-\cos^2 C)\iff$$  $$\iff a b\cos^2C +c^2\cos C-a b=0.$$ Solving this quadratic in $\cos C$ we have $$\cos C=(-c^2\pm \sqrt {c^4+4 a^2b^2}\;)/2 a b.$$ $$\text { But } \cos C=(a^2+b^2-c^2)/2 a b.$$ $$\text {Therefore }  \pm \sqrt {c^4+4 a^2b^2} =a^2+b^2$$ from which $c^4=(a^2+b^2)^2-4 a^2 b^2=(a^2-b^2)^2.$ So $$c^2=a^2-b^2 \lor c^2=b^2-a^2$$ and the triangle is right-angled.
