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