见解析
连接BC,AD,
根据直径所对的圆周角是直角,得∠C=∠D=90°,
根据相交弦定理,得AE?CE=DE?EB
∴AE?AC+BE?BD=AC2-AC?CE+BD2-BD?DE
=100-BC2+100-AD2-AC?CE-BD?DE
=200-BE2+CE2-AE2+DE2-AC?CE-BD?DE
=200+(DE+BE)(DE-BE)+(CE+AE)(CE-AE)-AC?CE-BD?DE
=200+BD(DE-BE)+AC(CE-AE)-AC?CE-BD?DE
=200-AE?AC-BE?BD,
∴AE?AC+BE?BD=100.