It is known that there exists a point $P$ within the interior of $\triangle ABC$ satisfying the following conditions: (i) $\angle PAB \ge 30^\circ$ and $\angle APB \ge \angle PCB + 30^\circ$; (ii) $BP \cdot BC=CP \cdot AB.$ Prove that $\angle BAC \ge 60^\circ$, and that equality holds only when $\triangle ABC$ is equilateral.