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In \(\triangle A B C, \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=\)
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Verified Answer
The correct answer is:
\(\frac{a^2+b^2+c^2}{\Delta^2}\)
\(\because\)\(r_1=\frac{\Delta}{s-a}\)
\(\therefore\)
\(\frac{1}{r_1}=\frac{s-a}{\Delta}\)
Similarly,
\(\frac{1}{r_2}=\frac{s-b}{\Delta} \frac{1}{r_3}=\frac{s-c}{\Delta}\)
and
\(\frac{1}{r}=\frac{s}{\Delta}\)
Now, \(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\)
\(\begin{aligned}
& =\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2} \\
& =\frac{s^2+a^2-2 a s+s^2+b^2-2 s b+s^2+c^2-2 s c+s^2}{\Delta^2} \\
& =\frac{4 s^2+a^2+b^2+c^2-2 s(a+b+c)}{\Delta^2} \\
& =\frac{4 s^2+a^2+b^2+c^2-4 s^2}{\Delta^2}=\frac{a^2+b^2+c^2}{\Delta^2}
\end{aligned}\)
\(\therefore\)
\(\frac{1}{r_1}=\frac{s-a}{\Delta}\)
Similarly,
\(\frac{1}{r_2}=\frac{s-b}{\Delta} \frac{1}{r_3}=\frac{s-c}{\Delta}\)
and
\(\frac{1}{r}=\frac{s}{\Delta}\)
Now, \(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\)
\(\begin{aligned}
& =\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2} \\
& =\frac{s^2+a^2-2 a s+s^2+b^2-2 s b+s^2+c^2-2 s c+s^2}{\Delta^2} \\
& =\frac{4 s^2+a^2+b^2+c^2-2 s(a+b+c)}{\Delta^2} \\
& =\frac{4 s^2+a^2+b^2+c^2-4 s^2}{\Delta^2}=\frac{a^2+b^2+c^2}{\Delta^2}
\end{aligned}\)
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