Question 66
Let $\triangle ABC$ be a triangle. Let $AD$, $BE$, $CF$ be cevians such that $\angle BAD = \angle CBE = \angle ACF$. Suppose these cevians concur at a point $\Omega$. (Such a point exists for each triangle and it is called a Brocard point.) Prove that
$$
\frac{A\Omega^2}{BC^2} + \frac{B\Omega^2}{CA^2} + \frac{C\Omega^2}{AB^2} \ge 1
$$
$$
\frac{A\Omega^2}{BC^2} + \frac{B\Omega^2}{CA^2} + \frac{C\Omega^2}{AB^2} \ge 1
$$