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
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?
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∠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
∠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
1年前
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