In $\triangle ABC$, the angle bisectors $BD$ and $CE$ intersect at $I$, with $D$ and $E$ on $AC$ and $AB$ respectively. The perpendicular from $I$ to $DE$ meets $DE$ at $P$. The extension of $PI$ meets $BC$ at $Q$. If $IQ = 2IP$, find $\angle A$.

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Let $\triangle ABC$ have sides $a, b, c$. By the angle bisector property,
$$
\frac{CD}{DA} = \frac{a}{c}
$$

Thus,
$$
\frac{CD}{b} = \frac{a}{a+c} \quad CD = \frac{ab}{a+c}
$$

Let $\angle IDE = \beta, \angle IED = \gamma$, then
$$
\beta + \gamma = 180^\circ - \angle D I E = 180^\circ - \angle B I C = \frac{B}{2} + \frac{C}{2}
$$

By the angle bisector property,
$$
\frac{ID}{IB} = \frac{CD}{CB} = \frac{\frac{ab}{a+c}}{a} = \frac{b}{a+c}
$$

By the sine definition and the sine rule,
$$
IP = ID \cdot \sin \beta = \frac{1}{2} IQ = \frac{1}{2} IB \times \frac{\sin \frac{B}{2}}{\sin \left(\frac{B}{2} + 90^\circ - \beta\right)}
$$

Combining (2), we get
$$
\frac{b}{a+c} \cdot 2 \sin \beta \sin \left(90^\circ - \beta + \frac{B}{2}\right) = \sin \frac{B}{2}
$$

Thus,
$$
\sin B \left(\cos \left(90^\circ - 2\beta + \frac{B}{2}\right) - \cos \left(90^\circ + \frac{B}{2}\right)\right) = (\sin A + \sin C) \sin \frac{B}{2}
$$

Canceling $\sin \frac{B}{2}$ on both sides, we get
$$
2 \cos \frac{B}{2} \left(\cos \left(90^\circ - 2\beta + \frac{B}{2}\right) - \cos \left(90^\circ + \frac{B}{2}\right)\right) = 2 \cos \frac{B}{2} \cos \frac{A-C}{2}
$$

Thus,
$$
\sin \left(2\beta - \frac{B}{2}\right) = \cos \frac{A-C}{2} - \cos \frac{A+C}{2} = 2 \sin \frac{A}{2} \sin \frac{C}{2} > 0
$$

Similarly,
$$
\sin \left(2\gamma - \frac{C}{2}\right) = 2 \sin \frac{A}{2} \sin \frac{B}{2} > 0
$$

If $A > 60^\circ$, then
$$
\begin{array}{l}
\sin \left(2\beta - \frac{B}{2}\right) > \sin \frac{C}{2} \\
\sin \left(2\gamma - \frac{C}{2}\right) > \sin \frac{B}{2}
\end{array}
$$

Thus,
$$
2\beta - \frac{B}{2} > \frac{C}{2} \quad 2\gamma - \frac{C}{2} > \frac{B}{2}
$$

Adding these, we get
$$
\beta + \gamma > \frac{B}{2} + \frac{C}{2}
$$

This contradicts (1). If $A < 60^\circ$, similarly (by reversing all inequalities) a contradiction arises.
Therefore, $A = 60^\circ$ (especially when $A = B = C = 60^\circ$, it is clear that $IQ = 2IP$).