7. As shown in Figure 2, construct two equilateral triangles $\triangle A B C$ and $\triangle C D E$ on the same side of line segment $A E$ $(\angle A C E<120^{\circ})$, and let $P$ and $M$ be the midpoints of line segments $B E$ and $A D$, respectively. Then $\triangle C P M$ is ( ).
(A) obtuse triangle
(B) right triangle
(C) equilateral triangle
(D) non-isosceles triangle
7.C.

It is easy to see that $\triangle A C D \cong \triangle B C E$.
Therefore, $\triangle B C E$ can be considered as $\triangle A C D$ rotated $60^{\circ}$ clockwise around point $C$. Since $M$ is the midpoint of segment $A D$ and $P$ is the midpoint of segment $B E$, $C P$ is obtained by rotating $C M$ $60^{\circ}$ clockwise around point $C$. Thus, $C P = C M$ and $\angle P C M = 60^{\circ}$. Therefore, $\triangle C P M$ is an equilateral triangle.