JEE Advance - Mathematics (1980 - No. 27)
$$ABC$$ is a triangle. $$D$$ is the middle point of $$BC$$. If $$AD$$ is perpendicular to $$AC$$, then prove that
$$$\cos A\,\cos C = {{2\left( {{c^2} - {a^2}} \right)} \over {3ac}}$$$
$\cos A \cos C = \frac{b^2 + c^2 - a^2}{2bc} \cdot \frac{a^2 + b^2 - c^2}{2ab} = \frac{2(c^2 - a^2)}{3ac}$
$\cos A \cos C = \frac{b^2 + c^2 - a^2}{2bc} \cdot \frac{a^2 + b^2 - c^2}{2ab} = \frac{b^4 - (a^2 - c^2)^2}{4ab^2c} = \frac{2(c^2 - a^2)}{3ac}$
Applying the Law of Cosines and the properties of medians to express \(\cos A\) and \(\cos C\), and using the given perpendicularity to simplify to the desired form. Considering Stewart's theorem.
Using the Law of Sines and the Law of Cosines along with the median properties will directly yield the solution.
This equation cannot be derived from the given conditions.