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In $\triangle A B C$, if the median $A D$ drawn through $A$ is perpendicular to the side $A C$, then $3 c a \cos A \cos C+2 a^2=$
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Verified Answer
The correct answer is:
$2c^2$
According to Apollonius Theorem
Now, $\quad 3 c a \cos A \cos C+2 a^2$
$$
\begin{aligned}
& =3 c a\left(\frac{b^2+c^2-a^2}{2 b c}\right)\left(\frac{a^2+b^2-c^2}{2 a b}\right)+2 a^2 \\
& =\frac{3}{4 b^2}\left[\left(b^2+c^2-3 b^2-c^2\right)\left(3 b^2+c^2+b^2-c^2\right)\right]+2 a^2 \\
& =\frac{3}{4 b^2}\left(-2 b^2\right)\left(4 b^2\right)+2 a^2=-6 b^2+2 a^2 \\
& =2\left(a^2-3 b^2\right)=2 c^2 \quad \text { [from Eq. (iii)] }
\end{aligned}
$$
Now, $\quad 3 c a \cos A \cos C+2 a^2$
$$
\begin{aligned}
& =3 c a\left(\frac{b^2+c^2-a^2}{2 b c}\right)\left(\frac{a^2+b^2-c^2}{2 a b}\right)+2 a^2 \\
& =\frac{3}{4 b^2}\left[\left(b^2+c^2-3 b^2-c^2\right)\left(3 b^2+c^2+b^2-c^2\right)\right]+2 a^2 \\
& =\frac{3}{4 b^2}\left(-2 b^2\right)\left(4 b^2\right)+2 a^2=-6 b^2+2 a^2 \\
& =2\left(a^2-3 b^2\right)=2 c^2 \quad \text { [from Eq. (iii)] }
\end{aligned}
$$
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