Example 29. Given an equilateral triangle $\mathrm{ABC}$, a line parallel to $\mathrm{AC}$ intersects $A B$ and $B C$ at $M$ and $P$, respectively. $D$ is the centroid of $\triangle P M B$, and $E$ is the midpoint of line segment $\mathrm{AP}$. Calculate the angles of triangle $\mathrm{DEC}$.
$$
\mathrm{B}\left(\mathrm{D},-60^{\circ}, \frac{1}{2}\right)
$$
K.
$$
\begin{array}{l}
P \xrightarrow{S\left(D,-60^{\circ}, \underset{2}{\frac{1}{2}}\right)} \mathrm{H}, \\
\mathrm{S} \xrightarrow{\mathrm{S}\left(\mathrm{D},-60^{\circ}, \frac{1}{2}\right)} \mathrm{R} \text {, } \\
\mathrm{S}\left(\mathrm{D},-60^{\circ}, \frac{1}{2}\right) \\
\therefore \mathrm{BP}-\rightarrow \mathrm{HK} \text {. } \\
\mathrm{HK} \text { is the midline of } \angle \mathrm{PMB} \text {. } \text { Therefore, } \mathrm{AP} \mathrm{F} \mathrm{E} \text {. } \\
\text { Extra, } \mathrm{SC}: \mathrm{BP} \\
=\mathrm{BA}: \mathrm{BM}=\mathrm{KE}: \mathrm{KH} \text {. } \\
S\left(D,-60^{\circ}, \frac{1}{2}\right) \\
\end{array}
$$

Moon r $\qquad$ $\rightarrow \mathrm{E}$.
That is, with $\mathrm{D}$ as the center, rotate $\mathrm{DC}$ by $60^{\circ}$, and reduce it to $\frac{1}{2}$ to become DE,
Therefore, $\angle \mathrm{EDC}=60^{\circ}, \mathrm{DE}=\frac{\mathrm{DC}}{2}$,
$\therefore \angle \mathrm{DEC}=90^{\circ}, \angle \mathrm{DCE}=30^{\circ}$.