如图,Rt△ABC中,∠BAC=90°,点D,E在BC上,且BE=AB,CD=AC,求∠DAC的度数

∵CA=CD
∴∠CAD=1/2(180°-∠C)=90°-1/2∠C
同理可得∠BAE=90°-1/2∠B
∵∠BAC =90°
∴∠B +∠C =90°
∴∠DAE =∠BAE+∠CAE-∠ABC
=90°-1/2∠C+90°-1/2∠B-90°
=90°-1/2(∠B+∠C)
=90°-45°
=45°