In $\triangle A B C$, if the median $A D$ drawn through $A$ is perpendicular to the side $A C$, then $3 c a…
- $c^2$
- $2c^2$
- $3c^2$
- $4c^2$
Solution

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} $
Asked in: AP EAMCET 2019 (20 Apr Shift 2)