小华爷
幼苗
共回答了20个问题采纳率:80% 举报
(1)在△B0C中∠BOC=180°-∠OBC-∠OCB=180°-∠B/2-∠C/2
=180°-(∠B+∠C)/2,
其中(∠B+∠C)/2=(180°-∠A)/2=90°-∠A/2,
所以∠BOC=180°-(90°-∠A/2)=90°+∠A.
(2)在△abc中,∠dbc=1/2*(∠a+∠c),∠bcd=1/2*(∠b+∠a)
∠d=180-1/2*(∠a+∠acb)-1/2*(∠abc+∠a)=180°-∠a-1/2∠acb-1/2∠abc
又∵∠abc+∠abc=180-∠a
∴∠d=180-1/2*(∠a+∠acb)-1/2*(∠abc+∠a)=180°-∠a-1/2∠acb-1/2∠abc=180°-∠a-1/2(180-∠a)=90-1/2∠a
(3)∵∠A+∠ABC+∠ACB=180
∴∠ABC+∠ACB=180-∠A
∵∠ACE=180-∠ACB,CD平分∠ACO
∴∠DCO=∠ACO/2=(180-∠ACB)/2=90-∠ACB/2
∵BD平分∠ABC
∴∠DBC=∠ABC/2
∵∠DCO是△PBC的外角
∴∠DCO=∠D+∠DBC=∠D+∠ABC/2
∴∠D+∠ABC/2=90-∠ACB/2
∴∠D=90-(∠ABC+∠ACB)/2=90-(180-∠A)/2=∠A/2
1年前
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