2x度∴∠C=∠E=20°∠BAC=∠DAE19、解:(1)图略20、解:∵AD为∠BAC的平分线,DE⊥AB于E,DF⊥AC于F(2)A1(-1,2);B1(-3,1);C1(2,-1)∴DE=DF(3)4.5∵S△ABC=S△ABD+S△ACD=∴=28∴DE=221、(1)22、(1)证明:∵BE⊥AC,CF⊥AB证明:∵∠ACB=90°,AD⊥CE,BE⊥CE∴∠BAC+∠ACF=∠CAB+∠ABE=90°∴∠BCE+∠DCA=∠DCA+CAD=90°,∴∠ACF=∠ABE∴∠BCE=∠CAD在△ABD和△GCA中在△BCE和△CAD中∴△ABD≌△GCA∴△BCE≌△CAD∴AD=AG解:(2)由(1)知△BCE≌△CAD解:(2)AD与AG的位置关系是:AD⊥AG∴AD=CE=5,BE=CD由(1)已证△ABD≌△GCA∵CD=CE-DE=5-3=2∴∠AGC=∠DAB∴BE=2㎝∵CF⊥AB,则∠AGF+∠GAF=90°∴∠DAB+∠GAF=90°即AD⊥AG23(1)证明:连接AC∵AE是BC边的中线,且AE⊥BC∴AG=AC∵AF是CD边的中线,且AF⊥CD∴AD=AC∴AB=AD(2)∠EAF=∠BAE+∠DAF证明:由已知得AE是BC的中垂线,AF是CD的中垂线∴AB=AC,AD=AC∴AE是∠BAC的平分线,AF是∠CAD的平分线∴∠BAE=∠CAE,∠CAF=∠DAF∴∠EAF=∠EAC+∠FAC=∠BAE+DAF24、证明:延长AD到F,使DF=AD,连接CF∵AD是BC边的中线∴BD=CD在△ABD和△FDC中F∴△ABD≌△FDC∴AB=CF,∠B=∠DCF∵AB=BC∴BC=CF,∠BAC=∠BCA∠ACF=∠ACB+∠DCF=∠BAC+∠B∵∠ACE=∠BAC+∠B∴∠ACF=∠ACE在△ACF和△ACE中∴△ACF≌△ACE∴∠1=∠2
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