第1个回答 2013-02-21
解:∵BO,CO分别平分∠ABC,∠ACB
∴∠1=1/2∠ABC,
∠2=1/2∠ACB
∵∠O=180°-(∠1=∠2)
∴∠O=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
则:∠O=90°+1/2∠A
2)解:∵BD,CD分别平分∠ABD,∠ACD
∴∠CBD=1/2∠ABC=∠A+∠ACB,
∠BCD=1/2∠ACB= ∠A+∠ABC
∵∠D=180°-(∠CBD-∠BDC)
∴∠D=180°-1/2(∠A+∠ACB+ ∠ABC+∠A)
=180°-1/2(180°+∠A)
=180°-90°-1/2∠A
=90°-1/2∠A
则:∠D=90°-1/2∠A
3)解:∵BD,CD分别平分∠ABC,∠ACE
∴∠ABC=2∠DBE,
∠ACE=2∠DCE=2(∠D+∠DBE)=2∠D+2∠DBE
∵∠A=∠ACE-∠ABC
∴∠A=2∠D+2∠DBE-2∠DBE
=2∠D
则:∠A=2∠D
第3个回答 2012-04-18
解:(1)∠BOC=180-∠OBC-∠OCB,∠A+∠ABC+∠ACB=180可以得到∠BOC=180-1/2(∠ABC+∠ACB)而∠ABC+∠ACB=180-∠A,∠BOC=180-1/2(180-∠A)=90+1/2∠A;
(2)图在哪?
(3)∠D+∠DBC+∠DCB=180,∠DCB=∠DCA+∠ACB,∠DCA=1/2∠ACE,∠ACE=∠ABC+∠BAC,∠DBC=1/2∠ABC,故∠D=180-∠DBC-∠DCB=180-(1/2∠ABC)-(∠DCA+∠ACB)=180-(1/2∠ABC)-(1/2∠ACE+∠ACB)=180-1/2∠ABC-1/2(∠ABC+∠BAC)-∠ACB=∠BAC-1/2∠BAC=1/2∠BAC,即∠D=1/2∠BAC,即∠A=2∠D;