如图,△ABC中,AD平分∠BAC,EF⊥AD交AB于点E,交AC于点F,交BC的延长线于点H.求证：∠H=2/1(∠A
如图,△ABC中,AD平分∠BAC,EF⊥AD交AB于点E,交AC于点F,交BC的延长线于点H.求证：∠H=2/1(∠ACB-∠B)
F改H,延长H到点E至直线AB
F改H,延长H到点E至直线AB
赞 xiongcuihua 幼苗
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∠H
= ∠AEH-∠B
= 90°-∠BAD-∠B
= 90°-(1/2)∠BAC-∠B
= 90°-(1/2)(180°-∠ACB-∠B)-∠B
= 90°-90°+(1/2)∠ACB+(1/2)∠B-∠B
= (1/2)(∠ACB-∠B)
= ∠AEH-∠B
= 90°-∠BAD-∠B
= 90°-(1/2)∠BAC-∠B
= 90°-(1/2)(180°-∠ACB-∠B)-∠B
= 90°-90°+(1/2)∠ACB+(1/2)∠B-∠B
= (1/2)(∠ACB-∠B)
1年前
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