▲ABC中,∠EAC=90°,AB=AC,点D为直线BC上一动点
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![▲ABC中,∠EAC=90°,AB=AC,点D为直线BC上一动点](/uploads/image/f/934871-23-1.jpg?t=%E2%96%B2ABC%E4%B8%AD%2C%E2%88%A0EAC%3D90%C2%B0%2CAB%3DAC%2C%E7%82%B9D%E4%B8%BA%E7%9B%B4%E7%BA%BFBC%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9)
∵∠EAD=∠EDA,∠EAD=∠CAD+∠EAC,∠EDA=∠B+∠FAD,∠FAD=∠CAD,∴∠EAC=∠B再问:为什么∠EAD=∠EDA呢?再答:FE平分∠AED,∠AEF=∠DEF,∠DFE
证明:∵AD∥BC,∴∠EAD=∠B,∠DAC=∠C.∵AD平分∠EAC,∴∠EAD=∠DAC.∴∠B=∠C.∴AB=AC.
∵角eac=30,ae⊥bc∴∠c=60∵∠b﹢∠bad=∠ade=80,∠bad=﹙180-∠b-∠c﹚÷2∴80-∠b=﹙180-∠b-60﹚÷2∴∠b=40
1∵∠EAF=25°AF是∠EAC的角平分线∴∠FAC=25°∵∠ACB=30°∴∠AFD=25°+30°=55°∵∠EAF=25°∴∠AEF=180°-25°-55°=100°∠AED=180°-1
(1)、证明:∵∠EAC=∠EDC,∠AFE=∠CFD ∴△AFE∽△DFC ∴∠C
证明:∵∠ABC=∠ACB∴∠EAC=∠ABC+∠ACB=2∠ABC∵AF平分∠EAC∴∠EAF=∠EAC/2=∠ABC∴AF∥BC
∵∠BAD=∠EAC∴∠BAC=∠EAD在△ABC和△AED中AB=AE∠BAC=∠EADAC=AD∴△ABC≌△AED(SAS)∴∠ABC=∠AED
设∠EAC=5x,则∠EAB=2x.DE是AB边上的垂直平分线,则∠EBA=2x∠BAC+∠CBA=90度=∠EAC+∠EAB+∠EBA=9xx=10度,∠B=2x=20度
证明:∵AB=AC,AD=AE∴∠B=∠C,∠ADE=∠AED(等边对等角)又∵∠ADE=∠B+∠BAD∠AED=∠C+∠CAE(三角形的一个外角等于与它不相邻的两内角之和)∴∠BAD=∠CAE(等量
应该是AB/AC=BD/BC吧?再问:那要怎么做?再答:证明:作CF∥AD,交AB于点F则∠EAD=∠AFC,∠DAC=∠ACF∵AD平分∠EAC∴∠EAD=∠DAC∴∠AFC=∠ACF∴AF=AC∵
证明:∵AD平分∠EAC,∴∠EAD=12∠EAC.又∵∠B=∠C,∠EAC=∠B+∠C,∴∠B=12∠EAC.∴∠EAD=∠B.所以AD∥BC.
∵∠DAB=∠EAC,∴∠DAB+∠BAE=∠EAC+∠BAE,即∠BAC=∠DAE,在ΔABC与ΔADE中:∠B=∠D,∠BAC=∠DAE,BC=DE,∴ΔABC≌ΔADE.只是需要全等吧.再问:半
ad的度数35再问:为什么再答:角cae等于角dab
∠DAE=65°又因为∠EAD为△ABD的外角所以∠EAD=∠B+∠D,所以∠D=65°-30°=35°
证明:点E在AD的垂直平分线上只需要证明AE=DC,只需要证明∠DAE=∠ADC,∠BAD+∠ABD=∠ADC∠BAD=DAC,∠ABD=∠CAE,所以∠ADC=∠ADE,得证啦~
∵DE垂直平分AB∴∠DAE=∠DBE∵∠AEC=∠DAE+∠DBE∵∠EAC:∠EAB=7:4∴∠CAE:∠EAB=7:8∵∠C=90°∴∠AEC=90°×8/15=48°肿么了吗
证明:过点E作AC的垂直平分线EF(图和二楼的一样)∵AD是△ABC的高∴∠ADB=∠ADE=90°在△ABD与△AED中∠BAD=∠DAE,AD=AD(公共边),∠ADB=∠AED∴△ABD≌△AE
证明:∵△ABD和△ACE都是等腰直角三角形,∴AB=AD,AE=AC,又∵∠BAD=∠CAE=90°,∴∠BAD+∠BAC=∠CAE+∠BAC,即:∠DAC=∠BAE,在△ABE和△ADC中,AB=
(1)、证明:∵∠EAC=∠EDC,∠AFE=∠CFD∴△AFE∽△DFC∴∠C=∠E又∵∠BAD=∠EAC∴∠BAD+∠DAC=∠EAC+∠DAC∴∠BAC=∠DAE又∵AC=AE∴△BAC≌△DA