解：（1）如图，延长$AE，$$DC$交于点$F。$
$\because AE$是$\angle BAD$的平分线，
$\therefore\angle BAE=\angle DAE。$
$\because AB// CD，$$\therefore\angle BAE=\angle F。$
$\therefore\angle DAE=\angle F。$$\therefore DA = DF。$
$\because E$是$BC$的中点，
$\therefore BE = CE。$
在$\triangle ABE$和$\triangle FCE$中，
$\begin{cases}\angle AEB=\angle FEC\\\angle BAE=\angle F\\BE = CE\end{cases}，$
$\therefore\triangle ABE\cong\triangle FCE。$
$\therefore AE = FE。$
又$\because DA = DF，$$\therefore\angle ADE=\angle FDE。$
$\therefore DE$是$\angle CDA$的平分线。
（2）由（1）知$\triangle ABE\cong\triangle FCE，$
$\therefore AB = FC = 5。$
$\because AD = DF = CD + CF = CD + 5，$$AD + 2CD = 10，$
$\therefore CD + 5+2CD = 10。$
$\therefore CD=\dfrac{5}{3}。$
