BE CE分别平分∠abc和∠acb.探索∠e与∠a的关系
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∵∠ACD=∠A+∠ABC,CP平分∠ACD∴∠PCD=∠ACD/2=(∠A+∠ABC)/2∵BP平分∠ABC∴∠PBC=∠ABC/2∴∠PCD=∠P+∠PBC=∠P+∠ABC/2∴∠P+∠ABC/2
(中间O点忘点了)如图所示,∠A=40度,所以∠ABC+∠ACB=180-40=140度因为bo和co分别平分∠abc和∠acb,所以∠OBC+∠OCB=140*1/2=70度所以∠BOC=180-7
∵BO、CO分别平分∠ABC和∠ACB,∴∠1=∠2,∠3=∠4.(1)∵∠A=60°,∴∠1+∠2+∠3+∠4=120°,∴∠1+∠4=60°,∴∠O=180°-60°=120°.(2)若∠A=10
根据三角形外角的性质,有∠ACD=∠A+∠ABC,∠PCD=∠P+∠PBC而,BP、CP分别是∠ABC、∠ACD的平分线,即有,∠PBC=(1/2)*∠ABC,∠PCD=(1/2)*∠ACD代入化简得
∵∠ABO=∠CBO,∠BCO=∠ACO,∴∠OBC+∠OCB=12(∠ABC+∠ACB)=180°−∠A2=180°−70°2=55°,∴在△BOC中,∠BOC=180°-55°=125°.
90+1/2`50根据三角形内角和知识,通过△ABC和△OBC进行等量代换得到的∵OC、OB平分∠ACB和∠ABC∴∠OCB=1/2∠ACB∠OBC=1/2∠ABC在△OBC中∠O=180°-(∠OC
∠BPC+∠PBC+∠PCB=180∠BPC+1/2∠ABC+1/2∠ACB=180(1)∠A+∠ABC+∠ACB=1801/2∠A+1/2∠ABC+1/2∠ACB=90(2)(1)—(2)得:∠BP
1.180°-(80°/2)-(60°/2)=110°2.180°-(180°-40°)/2=110°3.180°-(180°-n°)/2=90°+n°/2
∵∠A=∠ACE-∠ABC=2∠DCE-2∠DBC=2(∠DCE-∠DBC),∠D=∠DCE-∠DBC,∴∠A=2∠D=48°.
应该是证BE//DF证明:因为AD//BC ,根据平行线的性质,同旁内角互补且∠A=∠C所以∠ABC=∠CDA且四边形ABCD是平行四边形因为BE 、DF分别平分∠ABC和∠CDA
∵在△BOC中∠BOC+∠OBC+∠OCB=180∴∠OBC+∠OCB=180-110=70∵BO,CO分别平分∠ABC和∠ACB∴∠OBC=∠ABC/2∠OCB=∠ACB/2∴∠ABC/2+∠ACB
(1)四边形内角和=360度,又因为∠A=∠C=90°,所以∠ABC+∠ADC=360-90-90=180°又因为BE.DF分别平分∠ABC和∠ADC,所以∠1+∠2=二分之一*(∠ABC+∠ADC)
/>∵∠ACE=∠A+∠ABC,CD平分∠ACE∴∠DCE=∠ACE/2=(∠A+∠ABC)/2∵BD平分∠ABC∴∠DBC=∠ABC/2∴∠DEC=∠D+∠DBC=∠D+∠ABC/2∴∠D+∠ABC
.∵∠A=60°∴∠ABC+∠ACB=180°—∠A=120°∵BO,CO分别平分∠ABC和∠ACB∴∠OBC=1/2∠ABC,∠OCB=1/2∠ACB∴∠OBC+∠OCB=1/2(∠ABC+∠ACB
因为∠ACB=90,所以∠BAC+∠ABC=180-∠ACB=90因为PA、PB分别平分∠BAC和∠ABC,所以∠BAP+∠ABP=45所以∠APB=180-(∠BAP+∠ABP)=135
∠B+∠C=140°,均平分,则∠OBC+∠OCB=70°所以∠BOC=110°
证明:∵AD∥BC,∴∠A+∠ABC=180°,∠C+∠ADC=180°,∵∠A=∠C,∴∠ABC=∠ADC,∵BE、DF分别平分∠ABC和∠CDA,∴∠EBC=12∠ABC,∠EDF=12∠ADC,
∠ACB=90,由三角形的内角和为180,所以∠CAB+∠CBA=90PA、PB分别平分∠BAC和∠ABC,所以∠APB+∠ABP=90/2=45所以∠APB=180-45=135
根据三角形外角的性质,有∠ACD=∠A+∠ABC,∠PCD=∠P+∠PBC而,BP、CP分别是∠ABC、∠ACD的平分线,即有,∠PBC=(1/2)*∠ABC,∠PCD=(1/2)*∠ACD代入化简得
证明:因为AD平行BC,角A=角C所以四边形ABCD是平行四边形角ABC=角ADC又因为BE,DE分别平分角ABC和角CDA所以角ABE=角EBC=角ADF=角CDF(等量代换→角ABC=角ADC)所