[asy]
size(200);
pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0);
draw(arc(O,1, 0, 180));
EE=midpoint(arc(O,1, 0, 90));
draw(A--EE);
draw(A--D);
B=intersectionpoint(arc(O,1, 180, 0),EE--A);
draw(O--EE);
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt));
//Credit to TheMaskedMagician for the diagram
[/asy]

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

$\textbf{(A) }10^\circ\qquad\textbf{(B) }15^\circ\qquad\textbf{(C) }20^\circ\qquad\textbf{(D) }25^\circ\qquad\textbf{(E) }30^\circ$
1. **Draw the necessary lines and identify the given angles:**
   - Draw radius \( OB \).
   - We have two isosceles triangles, \( \triangle ABO \) and \( \triangle OBE \).
   - Given \( \angle EOD = 45^\circ \), then \( \angle EOA = 180^\circ - 45^\circ = 135^\circ \).

2. **Define the unknown angle:**
   - Let \( \angle BAO = x \).
   - Since \( \triangle ABO \) is isosceles with \( AB = OB \), we have \( \angle BOA = x \) and \( \angle ABO = 180^\circ - 2x \).

3. **Analyze triangle \( \triangle OBE \):**
   - Since \( \triangle OBE \) is isosceles with \( OB = OE \), we have \( \angle OEB = \angle OBE = 2x \).
   - Therefore, \( \angle BOE = 180^\circ - 4x \).

4. **Use the sum of angles around point \( O \):**
   - The sum of angles around point \( O \) is \( 360^\circ \).
   - We have \( \angle AOB + \angle EOB = 135^\circ \).
   - Substituting the expressions for these angles, we get:
     \[
     x + (180^\circ - 4x) = 135^\circ
     \]
     Simplifying this equation:
     \[
     x + 180^\circ - 4x = 135^\circ
     \]
     \[
     -3x + 180^\circ = 135^\circ
     \]
     \[
     -3x = 135^\circ - 180^\circ
     \]
     \[
     -3x = -45^\circ
     \]
     \[
     x = 15^\circ
     \]

5. **Conclusion:**
   - Therefore, \( \angle BAO = 15^\circ \).

The final answer is \( \boxed{15^\circ} \).