1. As shown in Figure 14, quadrilateral $A B C D$ is a square, and an equilateral triangle $\triangle A B E$ is constructed outward from side $A B$. $C E$ intersects $B D$ at point $F$. Then $\angle A F D=$ $\qquad$ degrees.
(Given that $\angle E C B=\angle C E B=$ $15^{\circ}$, and by the axial symmetry of the square, we get $\angle F A B=$ $\angle E C B=15^{\circ}$. Therefore, $\angle A F D=60^{\circ}$.)