18.98 As shown in the figure, in trapezoid $ABCD$, $AB$ and $CD$ are the two bases, and the diagonal $BD$ is equal in length to the leg $AD$. If $\angle DCB = 110^{\circ}$ and $\angle CBD = 30^{\circ}$, then $\angle ADB$ equals
(A) $80^{\circ}$.
(B) $90^{\circ}$.
(C) $100^{\circ}$.
(D) $110^{\circ}$.
(E) $120^{\circ}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] In $\triangle B D C$, $\angle B D C=180^{\circ}-110^{\circ}-$ $30^{\circ}=40^{\circ}$.

Since $D C / / A B$, we have $\angle B D C=\angle D B A=40^{\circ}$. $\triangle A B D$ is an isosceles triangle, so $\angle D A B=\angle D B A=40^{\circ}$.
Therefore, $\angle A D B=180^{\circ}-2 \cdot 40^{\circ}=100^{\circ}$.
Hence, the answer is $(C)$.