Question

如图3,在矩形ABCD中,AE垂直于BD,垂足为E,角DAE:角BAE=1:2,则角CAE=?

Answer 1

设AC与BD交点为O，角DAE=30度，角BAE=60度，那么角EBA=角OBA=90-60=30度，角BAO=角OBA=30度，角CAE=角BAE-角BAO=60-30=30度

Answer 2

∠DAE:∠BAE=1:2
∠BAD=90°
∴∠DAE=30°
∠ADB=60°
∴∠DAC=60°
∠CAE=30°