如图,△ABC中,∠C平分∠ACB,DE⊥BC于E,DF⊥AC于F.
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因为ABCD为平行四边形所以角ABC=角ADC因为BE平分∠ABCDF平分∠ADC所以角ABE=角EBC=角ADF=角FDC因为角BED=角A+角ABE角DFB=角C+角FDC所以角BED=角DFB因
∵BE平分∠ABC,DF平分∠ADC,∴∠ABE=∠EBC,∠ADF=∠FDC又∠ABC+∠ADC=360-90*2=180∴∠EBC+∠ADF=180/2=90又∠ABE+∠AEB=90,∠ABE=
证明:∵∠A+∠B+∠C+∠D=360°且∠A=∠C=90°∴∠B+∠D=180°又∠ABE=∠EBC,∠ADF=∠CDF∴∠EBC+∠CDF=90°又∠DFC+∠CDF=90°∴∠EBC=∠DFC∴
BE//DF证明:∵∠A=∠C=90º∴∠ABC+∠ADC=360º-∠A-∠C=180º∵BE平分∠ABC,DF平分∠ADC∴∠CBE=½∠ABC,∠CDF=
∵BD平分∠ABC∴∠ABD=1/2∠ABC∵∠A=∠ABD∴∠ABC=2∠A∵∠BDC=∠A+∠ABD∠C=∠BDC∴∠C=2∠A∵∠A+∠ABC+∠C=180°∴∠A+2∠A+2∠A=180°∠A
解;因为三角形的外角等于不相邻的两个内角之和,所以设∠ACB的外角为∠ACE,∠ACE=∠ABC+∠BAC.又因为BD平分∠ABC,所以∠DBC=1/2∠ABC同理:∠ACD=1/2∠ACE=1/2(
∠A=50°∴∠B+∠C=180-50=130°∠FBC+∠FCB=1/2(∠B+∠C)=65°∠BFC=180-(∠FBC+∠FCB)=180-65=115°
在.0是△ABC的旁心.相关证明利用两次角平分线性质定理就能推导出来,加油吧.
因为BC>BA,可在BC上取BE=BA,连接DE则⊿EBD≌⊿ABD,得ED=AD=DC,且∠BED=∠A,⊿DEC中,∠DEC=∠C,那么∠A+∠C=∠BED+∠DEC=180°.
证明:∵∠A+∠ABC+∠C+∠ADC=360,∠A=∠C=90∴∠ABC+∠ADC=360-(∠A+∠C)=180∵BE平分∠ABC∴∠ABE=∠ABC/2∴∠BED=∠A+∠ABE=90+∠ABC
∵在△ABC中,BI平分∠ABC,CI平分∠ACB,∠BIC=120°,∴∠ABC+∠ACB=2×(180°-120°)=120°,∴∠A=60°.故答案为:60°.
证明:在边BC上截取BE=BA,连接DE, &
∵EF垂直平分BD∴EF是BD的垂直平分线∴EB=ED,∵△BFE和△DFE是直角三角形,且EF=EF∴△BFE全等于△DFE(HL)∴∠EBF=∠EDF∵BD平分∠ABC∴∠ABD=∠CBD∴∠EB
∠A+2∠A+2∠A=180,所以∠A=36度再问:能不能详细一点,用因为所以再答:设角ABD=角DBC=x,,因为BD平分∠ABC,且∠A=二分之一∠ABC所以角A=x,角BDC=2x,角C=2x所
在△ABC中,∠ACE=∠A+∠ABC,在△DBC中,∠DCE=∠D+∠DBC,…(1)∵CD平分∠ACE,BD平分∠ABC,∴∠ACE=2∠DCE,∠ABC=2∠DBC,又∵∠ACE=∠A+∠ABC
证明:∵∠ABC=2∠C,BD平分∠ABC,∴∠ABD=∠DBC=∠C,∴BD=CD,在△ABD和△ACB中,∠A=∠A∠ABD=∠C,∴△ABD∽△ACB,∴ABAC=BDBC,即AB•BC=AC•
∵在△ABC中,∠A=50°,∴∠ABC+∠ACB=180°-50°=130°.∵BP平分∠ABC,CP平分∠ACB,∴∠PBC+∠PCB=12(∠ABC+∠ACB)=12×130°=65°,∴∠BP
∵BD平分∠ABC,CE平分∠ACB∴∠DBC=1/2∠ABC∠BCE=1/2∠ACB∴∠DBC+∠BCE=1/2(∠ABC+∠ACB)∵∠DBC+∠BCE=∠CED=65°∴∠ABC+∠ACB=65
/>115°60°70°2∠DEC+∠A=180°有疑问,