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In any
$$
\triangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2}
$$
equals to
Options:
$$
\triangle A B C, \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2}
$$
equals to
Solution:
2282 Upvotes
Verified Answer
The correct answer is:
$\sin ^2 A$
$\begin{aligned} & \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4 b^2 c^2} \\ & =\frac{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}{4 b^2 c^2} \\ & =\frac{16[s(s-a)(s-b)(s-c)]}{4 b^2 c^2} \quad(\because a+b+c=2 s) \\ & =\frac{4 \Delta^2}{b^2 c^2}=\left(\frac{2 \Delta}{b c}\right)^2=\sin ^2 A\left[\because \sin A=\frac{2 \Delta}{b c}\right]\end{aligned}$
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