一、(50 分) In $\triangle A B C$, the angle bisectors $B D$ and $C E$ intersect at $I$, with $D$ and $E$ on $A C$ and $A B$ respectively. The perpendicular from $I$ to $D E$ meets $D E$ at $P$. The extension of $P I$ meets $B C$ at $Q$. If $I Q=2 I P^{\prime}$, find $\angle A$.
Let $\triangle A B C$ have sides $a, b, c$. By the angle bisector property, we get $\frac{C D}{D A}=\frac{a}{c}$.
Thus, $\frac{C D}{b}=\frac{a}{a+c}, C D=\frac{a b}{a+c}$.
Let $\angle I D E=\beta, \angle I E D=\gamma$, 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, we get
$$
\frac{I D}{I B}=\frac{C D}{C B}=\frac{\frac{a b}{a+c}}{a}=\frac{b}{a+c}.
$$

Using the sine rule, we get $I P=I D \cdot \sin \beta=\frac{1}{2} I Q=\frac{1}{2} I B \times \frac{\sin \frac{B}{2}}{\sin \left(\frac{B}{2}+90^{\circ}-\beta\right)}$,
Combining with (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
$$
\sin \left(2 \beta-\frac{B}{2}\right)>\sin \frac{C}{2}, \sin \left(2 \gamma-\frac{C}{2}\right)>\sin \frac{B}{2},
$$

Thus,
$$
\begin{array}{c}
2 \beta-\frac{B}{2}>\frac{C}{2}, 2 \gamma-\frac{C}{2}>\frac{B}{2}, \\
\beta+\gamma>\frac{B}{2}+\frac{C}{2}.
\end{array}
$$

Adding these, we get a contradiction with (1).
If $A<60^{\circ}$, similarly (just change all the inequality signs), we also get a contradiction.
Therefore, $A=60^{\circ}$ (especially, when $A=B=C=60^{\circ}$, it is clear that $I Q=2 I P$).