解:
(1) 在$\mathrm{Rt}△ ABC$中,$∠ ACB=90°,$$∠ A=36°,$
$\therefore ∠ ABC=90° - 36°=54°。$
由旋转的性质可知 $BC=CE,$$∠ DCE=∠ ACB=90°,$$∠ E=∠ ABC=54°,$
$\therefore ∠ CBE=∠ E=54°,$
$\therefore ∠ BCE=180° - 54° -54°=72°,$
$\therefore ∠ BCD=∠ ACB - ∠ BCE=90° -72°=18°。$
(2) 设$AB$与$CD$的交点为$H,$连接$CM。$
由旋转的性质可知 $AC=CD,$$∠ CDE=∠ A=36°,$旋转角$α=∠ ACD=36°,$
$\therefore ∠ A=∠ ACD=36°,$
$\therefore AH=CH,$且$DE// AC,$
$\therefore ∠ A=∠ DMA=36°,$
又$\because ∠ CDE=∠ A=36°,$
$\therefore ∠ D=∠ DMA,$
$\therefore DH=HM,$
$\therefore AM=DC=AC,$
$\therefore ∠ AMC=∠ ACM=\frac{180° - 36°}{2}=72°,$
$\therefore ∠ DCM=∠ AMC - ∠ D=72° -36°=36°=∠ D,$
$\therefore DM=CM。$
$\because ∠ DCE=90°,$
$\therefore ∠ MCE=90° - ∠ DCM=54°,$
又$\because ∠ E=90° - ∠ D=54°,$
$\therefore ∠ MCE=∠ E,$
$\therefore ME=MC,$
$\therefore DM=ME,$即$M$是$DE$的中点。