9. In a convex quadrilateral $A B C D, \angle B A C=\angle C A D, \angle A B C=\angle A C D$, the extensions of $A D$ and $B C$ meet at $E$, and the extensions of $A B$ and $D C$ meet at $F$. Determine the value of
$$
\frac{A B \cdot D E}{B C \cdot C E} .
$$
9. Answer: 1 .

Since $\angle A C E$ is an exterior angle for the triangle $A B C$, we have $\angle A C E=\angle A B C+$ $\angle B A C$, hence $\angle A C E=\angle A C D+\angle C A D$. It follows that $\angle C A D=\angle D C E$, so the triangles $C E D$ and $A E C$ are similar. For this, we obtain $C E / A E=D E / C E$. But $\angle B A C=\angle C A E$, so we get $B C / C E=A B / A E$. Using these equalities, we get $(A B \cdot D E) /(B C \cdot C E)=1$.