如图,在△abc中,ad平分bc
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∵EF垂直平分AD∴EA=ED∴∠EAD=∠EDA∵AD平分角BAC,即∠BAD=∠CAD又∵∠EDA=∠B+∠BAD;∠EAD=∠CAE+∠CAD∴∠B=∠EDA-∠BAD=∠EAD-∠CAD=∠C
∵CD=DF∴∠DCF=∠DFC∵∠DFC=∠AFE∴∠DCF=∠AFE∵CE⊥AB∴∠AFE+∠BAD=90°∠EBC+∠DCF=90°∴∠BAD=∠EBC∴BD=AD
延长CD交AB于点E∵AD平分∠BAC∴∠BAD=∠CAD∵CD⊥AD∴∠ADE=ADC∵AD=AD∴⊿ADE≌⊿ADC﹙ASA﹚∴∠AED=∠ACD∵∠AED是△BCE的外角∴∠AED>∠B即∠AC
证明:延长CE交AB于F,∵CE⊥AD,∴∠AEC=∠AEF,∵AD平分∠BAC,∴∠FAE=∠CAE,在△FAE和△CAE中∵∠FAE=∠CAEAE=AE∠AEF=∠AEC,∴△FAE≌△CAE(A
(1)AD是BC的中垂线所以AB=AC,HB=HC,所以AB=AC,BD=CD,AD=AD三角形ABD全等于三角形ACD所以角BAD=角CAD所以评分啊(2)角BAD,CAD,ABH,ACH,HBD,
因为角EAD=角CAD,(AD平分角BAC)又:角EDA=角DAC,(DE//AC)所以,角EDA=角DAE又:EF垂直于AD所以,EF是AD的垂直平分线,∴FD=FA,(垂直平分线上的点到线段两个端
EF垂直平分AD所以AE=ED所以在三角形EAD中,∠EDA=∠EAD又∠EAD=∠EAC+∠CAD,∠EDC=∠B+∠DAB所以∠EAC+∠CAD=∠B+∠DAB又AD平分∠BAC所以∠DAB=∠C
证明:∵AD平分∠BAC,∴∠BAD=∠CAD,在△ABD和△ACD中AB=AC∠BAD=∠CADAD=AD,∴△ABD≌△ACD.
证明:∵∠B=90°-∠BAD∠C=90°-∠CAE-∠DAE∴∠B-∠C=∠CAE-∠BAD+∠DAE∵AE平分∠BAC∴∠CAE=∠BAE∴∠B-∠C=∠BAE-∠BAD+∠DAE∵∠BAE-∠B
∠EAC=180°-∠BAC;∠ACE=180°-∠BCA;∠ACE+∠EAC=180°-∠BAC+180°-∠BCA=360°-∠BCA-∠BAC;因为∠B=90°,故∠BCA+∠BAC=90°;所
如图∵EF垂直平分AD∴EA=ED∴∠EAD=∠EDA∵AD平分角BAC,即∠BAD=∠CAD又∵∠EDA=∠B+∠BAD; ∠EAD=∠CAE+∠CAD∴∠B=∠EDA-∠BAD=∠EAD
∵EF垂直平分AD∴EA=ED∴∠EAD=∠EDA∵AD平分角BAC,即∠BAD=∠CAD又∵∠EDA=∠B+∠BAD;∠EAD=∠CAE+∠CAD∴∠B=∠EDA-∠BAD=∠EAD-∠CAD=∠C
证明:∵AD平分∠EAC,∴∠EAD=12∠EAC.又∵∠B=∠C,∠EAC=∠B+∠C,∴∠B=12∠EAC.∴∠EAD=∠B.所以AD∥BC.
朋友这样做由三角形的正弦定律知sin∠AEB/AB=sin∠AEC/AC而AB>AC所以sin∠AEB>sin∠AEC因为AD平分∠BAC所以:∠ABE
设AB沿AD折叠点B落在AC上,这一点设为E,设BD=X,则AD=8-X,很容易证明:DE=BD=X,AE=AB=6,则由直角三角形的定理可知:AC=10=AE+CE则CE=4那么CE^2=16=CD
(1)因为角ABC=30°,角ACB=60°,所以角BAC=90°,又因为AE平分角BAC,所以角EAC=45°,AD⊥BC,所以角ADC=90°,角DAC=30°,那么角DAE=45°-30°=15
证明:作出AB边的高DE交AB于E∵AD=BD∴E为AB的中点,AB=2AE∵AB=2AC∴AE=AC∵AD平分∠BAC∴∠EAD=∠CAD又AE=AC,AD为公共边∴ΔEAD≌ΔCAD∴∠ACD=∠
过E分别作BA,BC,AC的垂线,交BA,BC,AC于M,N,P,∵BE平分∠ABC,∴△BEM≌△BEN(A,A,S)∴EM=EN.同理:EP=EN,∴EM=EP,即△AEM≌△AEP(H,L)∴∠
∵∠B=∠ADE-∠BAD=∠ADE-∠A/2 ∠CAE=∠DAE-∠DAC=∠DAE-∠A/2∵EF是AD的中垂线∴∠ADE=∠DAE∴∠B=∠CAE