Question

如图所示,在矩形ABCD中,对角线AC、BD交于点O,AE垂直BD交BD于点E,且∠BAE=3∠DAE,求∠CAE的度数

Answer 1

AD垂直AB,AE垂直BD,则∠DAE=∠ABE

∠BAE=3∠DAE,∠ABE+∠BAE=90,则∠ABE=90/4=22.5,∠BAE=3*22.5=67.5

AC、BD是矩形的对角线,则∠BAC=∠ABE=22.5

∠CAE=∠BAE-∠BAC=67.5-22.5=45