17. (10 points) As shown in Figure 4, $PA$ is a tangent to $\odot O$, with $A$ being the point of tangency. Line $PO$ intersects $\odot O$ at points $B$ and $C$. $D$ is the midpoint of $OC$. Connect $AD$ and extend it to intersect $\odot O$ at point $E$. If $PA = 2\sqrt{3}$ and $\angle APB = 30^\circ$, find
(1) the size of $\angle AEC$;
(2) the length of $AE$.
Three, 17. (1) Connect $A B$, it is easy to know, $\angle A P O=30^{\circ}$.
Thus, $\angle A O B=60^{\circ}$.
Also, $O A=O B$, so $\angle A B O=60^{\circ}$.
Therefore, $\angle A E C=\angle A B C=60^{\circ}$.
(2) From the given conditions, $A O=2$. Draw $A H \perp$ $B C$ at point $H$, then $A H=\sqrt{3}$.
In the right triangle $\triangle A H D$, $H D=2$.
Thus, $A D=\sqrt{3+2^{2}}=\sqrt{7}$.
By $B D \cdot D C=A D \cdot D E \Rightarrow D E=\frac{3 \sqrt{7}}{7}$.
Therefore, $A E=A D+D E=\frac{10 \sqrt{7}}{7}$.