ABC中,∠ABC,∠ACF的平分线BE,CF相交于点O,AG⊥BE于点G,
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1正确,因为∠ABC=∠ACB,∠EAC是三角形ABC的外角所以∠ACB=1/2∠EAC又因为AD平分∠EAC所以∠DAC=1/2∠EAC所以∠ACB=∠DAC所以AD平行BC2正确因为AD平行BC所
∠BAC+∠ACB=180°-∠B∠DAC+∠ACF=180°-∠BAC+180°-∠ACB=360°-(180°-∠B)=180°+∠B∠EAC+∠ACE=1/2(∠DAC+∠ACF)=90°+1/
证明:AD平分∠BAC,∴∠BAD=∠CADEF是垂直平分线∴∠FAE=∠FDE∴∠FAE-∠CAD=∠FDE-∠BAD∵∠FAE-∠CAD=∠CAF,∠FDE-∠BAD=∠ABC∴∠CAF=∠ABC
由题可知∠BAC+∠BCA=180°-42°=138°∴∠DAC+∠FCA=360°-138°=222°又∵E为∠DAC与∠ACF的角平分线的交点.∴∠CAE+∠ECA=222°/2=111°∴∠AE
证明:∵,△ABE、△ACF都是等边三角形∴∠EBA=∠FAC=90°FB=EBAC=FA∵AD⊥BC∴∠B+∠BAD=90°又∠BAC=90°∴∠B=∠DAC∴△BAD∽△ACD∠EBD=∠FAD∴
∵AD∥BC,∴∠ECB=∠FDA,又∠FAD=∠FDA,∴∠AFC=2∠FDA,又∵∠ACF=∠AFC,∴∠ACF=2∠FDA=2∠ECB,∴∠ACB=∠ACF+∠ECB=3∠ECB=75°,∴∠E
(1)2∠EAC=47°+∠BCA,(2)2∠ACE=47°+∠BAC,(1)+(2)得2(∠CAE+∠ACE)=180°-47°+94°=227°∠CAE+∠ACE=113.5°,∠AEC=180°
证明:∵△ABC是等边三角形,∴∠B=∠BAC=∠ACB=60°,AB=AC∵∠DAE=∠EAC+∠DAC=60°∠BAC=∠BAD+∠DAC=60°∴∠EAC=∠BAD∵∠ACF=180°-∠ACB
稍等再答:证明:∵AD平分∠BAC∴∠BAD=∠CAD∵EF垂直平分AD∴AF=DF∴∠FAD=∠FDA∵∠BAF=∠BAD+∠FAD,∠ACF=∠CAD+∠FDA∴∠BAF=∠ACF数学辅导团解答了
④是错误的,∠BDC=1/2∠ABC,∠ADB=1/2∠ABC,∵∠BAC≠∠ABC,∴∠ADB≠∠BDC,∴BD不是∠ADC的平分线.③∠DAC+∠DCA=1/2(∠EAC+∠ACF)=1/2(∠A
∵AD平分∠EAC,∴∠EAC=2∠EAD,∵∠EAC=∠ABC+∠ACB,∠ABC=∠ACB,∴∠EAD=∠ABC,∴AD∥BC,∴①正确;∵AD∥BC,∴∠ADB=∠DBC,∵BD平分∠ABC,∠
如图:作EM⊥BD、EN⊥BF、EO⊥AC垂足分别为M、N、O,∵AE、CE是∠DAC和∠ACF的平分线,∴EM=EO,EO=EN,∴EM=EN,∴BE是∠ABC的角平分线,∴∠ABE=12∠ABC=
角EBD=角DAF(EBA=CAF=60度,ABD+BAD=BAD+DAC,所以ABD=DAC,所以ABD+EBA=EBD=DAC+CAF=DAF)BD:AD=EB:FA(先证ABD与CBA相似,这个
根据三角形外角和公式可得∠A+∠ABC=∠ACF∠1+∠D=∠3∵BD,CD分别是∠ABC,∠ACF平分线∴∠ABC=2∠1=2∠2,∠ACF=2∠3=2∠4∴∠A=∠ACF-∠ABC=2∠3-2∠1
因为AD是△ABC中∠BAC的平分线所以∠BAD=∠DAC∠BAC=2∠BAD因为AD的垂直平分线EF交BC的延长线于F所以△ADF是等腰三角形∠DAF=∠ADF(等腰三角形的底角相等)∠ADF=∠B
∠A=77,∠B=58,∠C=45设∠BAD=∠CBE=∠ACF=x∵∠FDE=58∠DEF=45∴∠DFE=77则∠AFC=77-x∠ABD=58-x∠ECB=45-x∴∠A=77,∠B=58,∠C
∵FA平分∠DAC∴∠1=∠DAC/2∵FC平分∠ACF∴∠2=∠ACF/2∴∠1+∠2=(∠DAC+∠ACF)/2∵∠B+∠3+∠4=180° ∠B=48°∴∠3+∠4=132°∵∠3+∠
(1)证明:因为在三角形ABC中,角BAC=90度,AD垂直于BC于D,所以三角形ABD相似于三角形CAD,所以BD/AD=AB/AC,角ABD=角CAD,因为三角形ABE与三角形ACF是等边三角形,
∵AD⊥BC,BE⊥AC,∴∠ADC=∠BDF=∠BEA=90°,∴∠FBD+∠BFD=90°,∠DAC+∠AFE=90°,∵∠AFE=∠BFD,∴∠FBD=∠DAC=25°,在△BDF和△ADC中,