$20 \cdot 8$ In the right figure, $CD$ is the diameter of a semicircle with center $O$. Point $A$ is on the extension of $DC$; point $E$ is on the circumference of the semicircle, and point $B$ is the intersection of line segment $AE$ with the circumference of the semicircle (not coinciding with $E$). If the length of $AB$ equals the length of $OD$, and $\angle EOD$ is $45^{\circ}$, then $\angle BAO$ is
(A) $10^{\circ}$.
(B) $15^{\circ}$.
(C) $20^{\circ}$.
(D) $25^{\circ}$.
(E) $30^{\circ}$.
(30th American High School Mathematics Examination, 1979)
[Solution] Connect $BO$, and let $x$ and $y$ represent the measures of $\angle EOD$ and $\angle BAO$, respectively.

Notice that $AB = OD = OE = OB$. In $\triangle ABO$ and $\triangle AEO$, by the exterior angle theorem, we have
$$
\angle EBO = \angle BEO = 2y, \quad \text{and } x = 3y.
$$

Thus, $45^{\circ} = 3y$, which means $y = 15^{\circ}$. Therefore, the answer is $(B)$.