Question

已知△ABC,∠BAC为直角,线段BC上有E、D两点,连接AE和AD,∠BAD=∠BDA,∠CAE=∠CEA,求∠EAD

Answer 1

∠BAD=∠BDA=1/2(180-∠ABD)

∠CAE=∠CEA=1/2(180-∠ACE)

所以∠EAD=180-∠AED-∠ADE

=180-(180-∠CAE)-(180-∠BAD)

=∠CAE+∠BAD-180

=1/2(180-∠ABD)+1/2(180-∠ACE)-180

=1/2(∠ABD+∠ACE)

=1/2*90

=45度