18.18 Extend the sides $AD, BC$ of the convex quadrilateral $ABCD$ through $D, C$ to meet at point $E$. Let $s=\angle CDE + \angle DCE$, and $s'=\angle BAD + \angle ABC$. If $r=\frac{s}{s'}$, then
(A) either $r=1$, or $r>1$.
(B) either $r=1$, or $r1$.
(E) $r=1$.
(19th American High School Mathematics Examination, 1968)
[Solution] In $\triangle EDC$, we have
$$
\angle E + \angle CDE + \angle DCE = \angle E + s = 180^{\circ}.
$$

In $\triangle EAB$, we also have
$$
\angle E + \angle BAE + \angle ABE = \angle E + s' = 180^{\circ}.
$$

Thus, $s = s' = 180^{\circ} - \angle E$.
Therefore, the answer is $(E)$.