Question

如图所示,△ABC中,∠ACB＞90°,AE平分∠BAC,AD垂直于BC交BC的延长线于点D,则∠DAE,∠ACB,∠B

Answer 1

∠DAE=1/2(∠ACB-∠B)

∠ADB=90度

∠ACB=∠ADB+∠DAC

所以∠ACB=90-∠DAC

因为∠ADB+∠DAB+∠B=180度

∠DAB=∠DAE+∠BAE

所以∠DAE+∠BAE+∠B=90度

因为AE平分∠BAC

所以∠CAE=∠BAE

∠CAE+∠DAC=∠DAE

所以∠BAE=∠DAE-∠DAC

所以∠ACB=∠DAE+∠DAE-∠DAC+∠B+∠DAC

∠ACB=2∠DAE+∠B

所以∠DAE=1/2(∠ACB-∠B)