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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.

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.

Grade:11

1 Answers

mycroft holmes
272 Points
5 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

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