【答案】$ABC$;$ACB$;$DBC$;$ECB$;$DBF$;$F$;$ECB$;同位角相等,两直线平行【解析】$\because BD$平分$\angle ABC$,$CE$平分$\angle ACB$(已知)$\therefore \angle DBC=\dfrac{1}{2}\angle ABC$,$\angle ECB=\dfrac{1}{2}\angle ACB$(角平分线的定义)又$\because \angle ABC=\angle ACB...
{\circ }$,$\therefore \angle BOE=\angle OBF$,$\therefore EC\ykparallel BF$;$\because \angle ABC=\angle ACB$,$\because BD$平分$\angle ABC$,$CE$平分$\angle ACB$,$\therefore \angle DBC=\dfrac{1}{2}\angle ABC$,$\angle ECB=\dfrac{1}{2}\angle ACB$,$\therefore \angle DBC=\...
(3)根据角平分线性质可得∠FBC= 1 2 ∠PBC,∠FCB= 1 2 ∠PCB,根据三角形内角和为180°可得∠ABC+∠ACB=180°-∠A,即可解题. 解答: 解:(1)∵BD平分∠ABC,CD平分∠ACB, ∴∠DBC+∠DCB= 1 2 (∠ABC+∠ACB), ∵∠ABC+∠ACB=180°-∠A, ∴∠DBC+∠DCB= 1 2 (180°-∠A), ∴∠D=180°...
解答 解:∵BD平分∠ABC,CE平分∠ACB ( 已知 )∴DBC=1212∠ABC,∠ECB=1212∠ACB,∵∠ABC=∠ACB (已知)∴∠DBC=∠ECB.∠DBF=∠F(已知)∴∠F=∠ECB∴EC∥DF(同位角相等两直线平行).故答案为:ABC;ACB;DBC;ECB;DBF;F;ECB;同位角相等两直线平行. 点评 此题考查了平行线的判定,熟记同位角相等,两直线平行...
(1)∵ BD平分∠ ABC,CD平分∠ ACB,∴∠ DBC+∠ DCB=1/2(∠ ABC+∠ ACB),∵∠ ABC+∠ ACB=180°-∠ A,∴∠ DBC+∠ DCB=1/2(180°-∠ A),∴∠ D=180°-(∠ DBC+∠ DCB)=90°+1/2∠ A;(2)∵ BE平分∠ ABC,CE平分∠ ACM,∴∠ ACE=1/2∠ ACM,∠ CBE=1/2∠ ABC,∵∠ ACM...
如图,在△ABC中,BD、CD分别平分∠ABC、∠ACB,点M、N、Q分别在AB、AC、BC的延长线上,BE、CE分别平分∠MBC、∠NCB.(1)若∠A=60°,①∠B
15.如图.∠ABC=∠ACB.BD.CD.BE分别平分△ABC的内角∠ABC.外角∠ACP.外角∠MBC.以下结论:①AD∥BC,②DB⊥BE,③∠BDC+∠ABC=90°,④∠A+2∠BEC=180°,⑤DB平分∠ADC.其中正确的结论有:①②③④
解:解:(1).BD、CD是∠ABC和∠ACB的角平分线∠DBC= ∠ABC, ∠DCB= ∠ACB,∵∠ABC+∠ACB=180°-∠A,∠BDC=180°-∠DBC-∠DCB=180°∠ABC+∠ACB)=180°-(180°-∠A)=90∠A,∴∠BDC=90°+ ∠A.2).∠ACM是△ABC的外角∴∠ACM=∠A+∠ABC,.BE平分∠ABC,CE平分外角∠ACM,∠EBC=∠ABE= ∠...
(1)∵BD平分∠ABC,CD平分∠ACB,∴∠DBC+∠DCB= 1 2(∠ABC+∠ACB),∵∠ABC+∠ACB=180°-∠A,∴∠DBC+∠DCB= 1 2(180°-∠A),∴∠D=180°-(∠DBC+∠DCB)=90°+ 1 2∠A;(2)∵BE平分∠ABC,CE平分∠ACM,∴∠ACE= 1 2∠ACM,∠CBE= 1 2∠ABC,∵∠ACM=∠A+∠ABC,∠A+∠ABC+∠ACB=...
18.如图.已知∠ABC=∠ACB.BD平分∠ABC.CE平分∠ACB.F是BC延长线上一点.且∠DBC=∠F.求证:EC∥DF.