Let $ABCD$ be a square. Consider a variable point $P$ inside the square for which $\angle BAP \ge 60^\circ.$ Let $Q$ be the intersection of the line $AD$ and the perpendicular to $BP$ in $P$. Let $R$ be the intersection of the line $BQ$ and the perpendicular to $BP$ from $C$.

[list]
[*] [b](a)[/b] Prove that $|BP|\ge |BR|$

[*] [b](b)[/b] For which point(s) $P$ does the inequality in [b](a)[/b] become an equality?[/list]
### Part (a): Prove that \( |BP| \ge |BR| \)

1. **Setup and Definitions:**
   - Let \(ABCD\) be a square with side length 1.
   - Let \(P\) be a point inside the square such that \(\angle BAP \ge 60^\circ\).
   - Let \(Q\) be the intersection of line \(AD\) and the perpendicular to \(BP\) at \(P\).
   - Let \(R\) be the intersection of line \(BQ\) and the perpendicular to \(BP\) from \(C\).
   - Let \(\alpha = \angle BAP\) and \(\beta = \angle ABP\).

2. **Cyclic Quadrilateral:**
   - Since \(\angle BAP = \alpha\) and \(\angle BQP = \alpha\), quadrilateral \(ABPQ\) is cyclic.
   - Therefore, \(\angle PBQ = 90^\circ - \alpha\) and \(\angle ABQ = \alpha + \beta - 90^\circ\).

3. **Trigonometric Relationships:**
   - Using the Law of Sines in \(\triangle BPQ\):
     \[
     BP = BQ \sin \alpha
     \]
   - Using the Law of Sines in \(\triangle BRX\) (where \(X\) is the intersection of \(\overline{CR}\) and \(\overline{BP}\)):
     \[
     BR = \frac{BX}{\sin \alpha} = \frac{\sin \beta}{\sin \alpha}
     \]

4. **Inequality to Prove:**
   - We need to show:
     \[
     BP \ge BR \implies BQ \sin \alpha \ge \frac{\sin \beta}{\sin \alpha}
     \]
   - Simplifying, we get:
     \[
     BQ \sin^2 \alpha \ge \sin \beta
     \]

5. **Maximum Value of \(\sin \beta \sin (180^\circ - \alpha - \beta)\):**
   - The maximum value of \(\sin m \sin (n - m)\) is attained when \(m = \frac{n}{2}\).
   - Therefore:
     \[
     \sin \beta \sin (180^\circ - \alpha - \beta) \le \sin^2 \left(90^\circ - \frac{\alpha}{2}\right)
     \]
   - Since \(90^\circ - \frac{\alpha}{2} \le 60^\circ\), we have:
     \[
     \sin^2 \left(90^\circ - \frac{\alpha}{2}\right) \le \sin^2 \alpha
     \]

6. **Conclusion:**
   - Thus, we have:
     \[
     \sin \beta \sin (180^\circ - \alpha - \beta) \le \sin^2 \alpha
     \]
   - This implies:
     \[
     BP \ge BR
     \]

### Part (b): For which point(s) \(P\) does the inequality in (a) become an equality?

1. **Equality Condition:**
   - Equality holds when \(\alpha = 60^\circ\) and \(\beta = 60^\circ\), or when \(\triangle ABP\) is equilateral.

2. **Conclusion:**
   - Therefore, the inequality \( |BP| \ge |BR| \) becomes an equality when \(P\) is such that \(\triangle ABP\) is equilateral.

The final answer is \( \boxed{ P } \) such that \(\triangle ABP\) is equilateral.