Question

如图,已知矩形ABCD中,∠EDC=15°,AD=2AB.求证：AD=AE

Answer 1

tan15°=tan(45°-30°)=(tan45°-tan30°)/(1+tan45°tan30°)=2-根号3

设AB=DC=m,则AD=BC=2m

∵∠EDC=15°

∴EC=DCtan15°=mtan15°

BE=BC-EC=2m-mtan15°

AE^2=AB^2+BE^2

=m^2+(2m-mtan15°)^2

=m^2{1+(2-tan15°)^2}

=m^2{1+(2-2+根号3)^2}

=4m^2

∴AE=2m

∴AD=AE