Let $A B C D$ be a square. Consider a variable point $P$ inside the square for which $\angle B A P \geq 60^{\circ}$. Let $Q$ be the intersection of the line $A D$ and the perpendicular to $B P$ in $P$. Let $R$ be the intersection of the line $B Q$ and the perpendicular to $B P$ from $C$. (a) Prove that $|B P| \geq|B R|$. (b) For which point(s) $P$ does the inequality in (a) become an equality?