In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$. Point $A$ lies on the extension of $DC$ past $C$; 
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle. 
If length $AB$ equals length $OD$, and the measure of $\measuredangle EOD$ is $45^\circ$, then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$
Solution by e_power_pi_times_i
Because $AB = OD$, triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$. Then $\measuredangle ABO = 180^\circ-2\theta$, and $\measuredangle EBO = \measuredangle OEB = 2\theta$, so $\measuredangle BOE = 180^\circ-4\theta$. Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$. Therefore $\theta+180-4\theta = 135^\circ$, and $\theta = \boxed{\textbf{(B) } 15^\circ}$.