1. If all the diagonals of a convex $n$-gon $F(n \geqslant 4)$ are equal, then
(A) $F \in\{$ quadrilateral $\}$;
(B) $F \in\{$ pentagon $\}$;
(C) $F \in\{$ quadrilateral $\} \cup\{$ pentagon $\}$;
(D) $F \in\{$ equilateral polygon $\} \cup\{$ equiangular polygon $\}$.
1. (C)

Using a regular pentagon as a counterexample can refute conclusion (A); using a square as a counterexample can refute conclusion (B); using an isosceles trapezoid as a counterexample can refute conclusion (D). Therefore, the correct conclusion can only be (C). In fact, to affirm (C) positively, it is still necessary to prove that $nA_{1} A_{3} + A_{1} A_{n}$. This leads to a contradiction.
Thus, the number of sides of a convex polygon with all diagonals equal cannot be $\geqslant 6$, i.e., it can only be a quadrilateral or a pentagon. Examples of a square and a regular pentagon show that convex quadrilaterals and pentagons with all diagonals equal do exist.