$17 \cdot 30$ In $\triangle A D E$, $\angle A D E=140^{\circ}$, point $B$ and point $C$ are on sides $A D$ and $A E$ respectively. If $A B, B C, C D$ and $D E$ are all equal in length, then the degree measure of $\angle E A D$ is
(A) $5^{\circ}$.
(B) $6^{\circ}$.
(C) $7.5^{\circ}$.
(D) $8^{\circ}$.
(E) $10^{\circ}$.
(29th American High School Mathematics Examination, 1978)
[Solution] Let $x=\angle B A C=\angle B C A, y=\angle C B D=\angle C D B, \approx z=$ $\angle D C E=\angle D E C$.

In $\triangle A B C, \triangle A C D$ and $\triangle A D E$, by the properties of the sum of interior and exterior angles of a triangle, we get
$$
\begin{array}{l}
y=2 x, \quad z=x+y=3 x, \\
x+\angle A D E+z=180^{\circ},
\end{array}
$$

So $140^{\circ}+4 x=180^{\circ}$,
we get $x=10^{\circ}$.
Therefore, the answer is $(E)$.