$解：∵CD是⊙O的直径，$
$∴∠CBD=90°，$
$∴S_{△CBD}=\frac {1}{2}×CD×OB=\frac {1}{2}×2R×R=R^2，$
$∵⊙O的半径为R，AB⊥CD，$
$∴BC=BD=\sqrt{2}R，$
$∵CD是⊙O的直径，$
$∴∠CBD=90°，$
$∴S_{扇形CBD}=\frac {1}{2}×\frac {π}{2}×BC^2$
$=\frac {π}{2}R^2，$
$∴S_{阴影ACED}= S_{半圆ACD}-S_{弓形}= S_{半圆ACD}-(S_{扇形CBD}-S_{△CBD})$
$=\frac {1}{2}πR^2-(\frac {π}{2}R^2-R^2)$
$=R^2.$
