$18 A B C D$ is a square, $E, F$ are the midpoints of $B C, C D$ respectively. By folding $A E, A F, E F$ to form a tetrahedron, points $C, B, D$ coincide, denoted as $P$, then the dihedral angle between plane $P E F$ and $A E F$ is $\qquad$
$18 \arcsin \frac{2 \sqrt{2}}{3}$. Let $A C \cap E F=G$, connect $P G$, we know that $\angle P G A$ is the plane angle of the dihedral angle to be found. By the problem, $A P \perp P E, A P \perp P F$, thus $A P \perp$ plane $P E F$, hence $A P \perp P G$. Let the side length of the square be $a$, then $P A=a, G A=\frac{3 \sqrt{2}}{4} a$, in the right triangle $\triangle P G A$, $\sin \angle P G A=\frac{P A}{G A}=$ $\frac{2 \sqrt{2}}{3}, \angle P G A=\arcsin \frac{2 \sqrt{2}}{3}$.