如图10,AB=AF,BA=EF,角B=角F,D是CE的中点,求证AD垂直CE

∵ACDE是平行四边形,∴CF=FE,AF=DF．∴S△AEF=S△CDF=S△CAF=1/2S△ACD=1/4S▱ABCD．∵S平行四边形ABCD=12,∴S△AEF=3．

∵DE⊥BC∴∠FDB=∠EDC∴∠B=90°—∠F∠C=90°—∠CED∵AE=AF∴∠F=∠AEF∵∠AEF=∠CED∴∠F=∠CED∴∠B=∠C∴AB=AC

DE∥AF∵AB=AC,AF⊥BC∴∠B=∠C,AF是∠BAC的角平分线∴∠FAC=(180°—∠B—∠C)/2∵AD=AE∴∠ADE=∠AED∵∠EAD=∠B+∠C∴∠AED=(180°—∠B—∠C

/>∵AB=AC∴

证明:∵DE⊥AB,DF⊥AC∴∠AED=∠AFD=90∵AE=AF,AD=AD∴△AED≌△AFD(HL)∴DE=DF再答：∴角1=角2再答：AD平分角BAC再答：亲做完啰麻烦采纳哦

因为AE=AF 则∠AEF=∠AFE因为∠AFE=∠DFC ∠DFC+∠DCF=90°所以∠BCA+∠BED=90°因为∠ABC+∠BED=90°则∠ABC=∠ACB所以AB=AC

很明显的平行关系嘛~AB=AC,AF⊥BC所以AF是

∠BAF=∠CAF,∠AED=∠ADE.又∠BAF+∠CAF+∠CAD＝180°＝∠AED+∠ADE+∠CAD∴=∠CAF=∠AED,DE‖AF

∵AB=AC∴

（1）图中是通过绕点A旋转90°，使△ABE变到△ADF的位置．证明：（2）BE=DF，BE⊥DF；延长BE交DF于G；由△ABE≌△ADF，得BE=DF，∠ABE=∠ADF；又∠AEB=∠DEG；∴

证明：∵四边形ABCD是平行四边形，∴AB=CD，AB∥CD，∴△AFE∽△DCE，∵AEDE=AFCD，∵AB=AF，AB=CD，∴AF=CD，∴AE=DE．

DF⊥BC∴∠CED+∠C=90º,∠F+∠B=90ºAB=AC∴∠C=∠B∴∠F=∠CED又∠CED=∠AEF∴∠F=∠AEF∴AE=AF

∵ACDE是平行四边形,∴CF=FE,AF=DF．∴S△AEF=S△CDF=S△CAF=S△ACD=S▱ABCD．∵S平行四边形ABCD=12,∴S△AEF=3．

（1）全等证明：∵四边形ABCD是正方形∴AD=AB,DA⊥AB∴∠DAF=∠DAB=90°∵AF=1/2AB∴AF=1/2AD∵E是AD中点∴AE=DE∴AF=AE∵AD=AB,∠DAF=∠DAB∴

证明:BE=DF∵E是AD的中点AF=1/2AB且在正方形ABCD中∴AF=AEAD=AB∵△ABE≌△ADF∴BE=DF

过A点作AD⊥BC,垂足为D.则AD为∠BAC的角平分线∵AE=AF∴∠E=∠AFE∵∠BAC=∠E+∠AFE∴∠E=1/2∠BAC∴∠E=∠BAD∴EF∥AD∵AD⊥BC,EF∥AD∴EF⊥BC

额,我也很想帮你,可是图在哪里呢.我单靠你的文字表述实在不知道图是怎么样的.你把图传上来,再追问我,我会帮你回答的.再问：这儿再答：∠‖⊥∵∵AB=AC∴∠B=∠C∵DE⊥BC∴∠EDB=∠EDC=9

∵AE=AF，∴∠F=∠AEF（等边对等角）．又∵∠AEF=∠CED（对顶角相等），FD⊥BC，∴∠F+∠B=90°，∠C+∠CED=∠C+∠F=90°，∴∠B=∠C（等量代换），∴AB=AC（等角对

证明：∵AE＝AF∴∠F＝∠AEF∵∠CED＝∠AEF∴∠F＝∠CED∵DE⊥BC∴∠B+∠F＝90,∠C+∠CED＝90∴∠C+∠F＝90∴∠B＝∠C∴AB＝AC

证明：连接AC、DF．∵ABCD是平行四边形,∴AB=CD,AB∥CD．∵AF=AB,∴AF=CD,且AF∥CD,∴ACDF是平行四边形．∵E是平行四边形ACDF对角线交点∴E是AD中点祝楼主学习进步