# In a triangle (a+b+c) (a+b-c) (b+c-a) (c+a-b) =8a^2b^2c^2/ a^2+b^2+c^2 then the triangle is right angled prove that.

mycroft holmes
272 Points
6 years ago
Let s = (a+b+c)/2. Then LHS = 16 s(s-a)(s-b)(s-c) = 16A2 = 4a2b2 sin2C.

Hence we have $4a^2b^2 \sin^2 C = \frac{{}8a^2b^2c^2}{a^2+b^2+c^2}$

or $\sin^2 C = \frac{2c^2}{a^2+b^2+c^2}$ and hence

$\cos^2 C = \frac{a^2+b^2-c^2}{a^2+b^2+c^2}$. Using the cosine formula, this implies

$\frac{(a^2+b^2-c^2)^2}{4a^2b^2} = \frac{a^2+b^2-c^2}{a^2+b^2+c^2}$ and hence

$(a^2+b^2-c^2)(a^2+b^2+c^2)= 4a^2b^2$

or $(a^2+b^2)^2 - c^4= 4a^2b^2$

$\Rightarrow (a^2+b^2)^2 - 4a^2b^2- c^4= 0$ or
$\Rightarrow (a^2-b^2)^2 - c^4= 0$
$\Rightarrow (a^2-b^2+c^2) (a^2-b^2- c^2)= 0$
$\Rightarrow a^2+c^2 = b^2$ or $\Rightarrow a^2 = b^2+c^2$ and hence we have a right angled triangle