1. As shown in Figure 1, in the equilateral $\triangle ABC$, points $D$ and $E$ are on sides $BC$ and $CA$ respectively, such that $CD = AE$, $AD$ intersects $BE$ at point $P$, and $BQ \perp AD$ at point $Q$. Then $\frac{QP}{QB}=$ $\qquad$
1. $\frac{\sqrt{3}}{3}$.

It is easy to see that $\triangle A B E \cong \triangle C A D$. Then
$$
\begin{array}{l}
\angle A B E=\angle C A D \\
\angle D P B=\angle P A B+\angle A B E \\
=\angle P A B+\angle C A D=60^{\circ}
\end{array}
$$

Therefore, $\frac{Q P}{Q B}=\cot 60^{\circ}=\frac{\sqrt{3}}{3}$.