ab=ac,bf=cf,求证角B=角C

AB=CD,AE=CF,AE-EF=CF-EF,AF=CE,<AFB=<CED=90度,RT△AFB≌RT△CED,BF=DE,《BFG=〈DEG=90度,〈BGF=〈DGE,（对顶角相等

AB=CD,AE=CF,AE-EF=CF-EF,AF=CE,

∵BF⊥ACCE⊥AB∴∠BED=∠AED=∠CFD=∠AFD∵∠EDB=∠CDF∠BED=∠CFDBE=CF∴△BED≌△CFD∴DE=DF∵DE=DFAD=AD∠AED=∠AFD∴△AED≌△AF

证明：因为已知AB∥CD,那么,∠A=∠C,又已知DE⊥AC,BF⊥AC,那么,∠AFB=∠CED,则,∠B=∠D那么,在三角形ABF和三角形CED中,∠A==∠C,AB=CD,∠B=∠D所以,三角形

∵BE=CF∴BE+EC=CF+EC即BC=EF∵AB=DE,AC=DF∴△ABC≌△DEF（S.S.S）∵∠B=∠DEF,∠ACB=∠F∴AB∥DE,AC∥DF

(1)∵DE⊥AC,BF⊥AC∴∠DEC=∠BFA=Rt∠又AB=AC,DE=BF∴Rt△DEC≌Rt△BFA(HL)∴AF=CE∴AF-EF=CE-EF即AE=CF(2)∵Rt△DEC≌Rt△BFA

证明：连接BC∵AB＝AC∴点A在BC的垂直平分线上∵DB＝DC∴点D在BC的垂直平分线上∴AD垂直平分BC∴BF＝CF再问：-10(2/7)×9(5/7)[-999（2/3)]^2老师这两题计算怎么

几年级的因为BE=CF∠BDE=∠CDF对顶角相等∠DFC=∠DEB因为垂直所以△DEB与△DFC全等角角边所以DF=DE所以AD平分∠BAC

根据全等三角形的判断：直角三角形斜边和一条直角边(HL)△AFB与△CED全等所以AF=CE又EF=FEAF-EF=AE=CE-FE=CF所以AE=CF又△AFB与△CED全等,角DCE=角BAF根据

嗯,一样的!这里我也回答下,行吗?在△ABD与△ACD中AB=ACBD=DCAD=AD∴△ABD≌△ACD∴∠ADB=∠ADC∴∠BDF=∠FDC在△BDF与△FDC中BD=DC∠BDF=∠FDCDF

连接PC∵AB=AC,AD是BC边上的中线∴∠BAP=∠CAP∴△BAP≌△CAP∴PB=PC,∠ABP=∠ACP∵CF‖AB∴∠F=∠ABP∴∠F=∠ACP∵∠EPC=∠CPF∴△EPC∽△CPF∴

应该是“BE=CF"9BE=CF∴BE+EC=EC+CF即BC=EF∵AB=DEAC=DF∴△ABC≌△DEF∴∠ABC=∠DEF∠ACB=∠DFE∴AB∥DEAC∥DF

基本解题思路：1.连接AF,EF与AB交与H,连接DH并延长与AF交与G.2,三角形AGD与三角形DBA全等,AB=DG,FG=FB3,DG平行AC4,FG/FA=Dg/AC.FB/FD=AB/AC5

∵AB∥DC.∴∠DCE=∠BAF.∵DE⊥AC,BF⊥AC.∴∠DEA=90°=∠BFC.∵AE=CF.∴AE+EF=CF+EF.即AF=CE.∴△CDE≌△ABF（ASA）∴DE=BF.

证明：∵AD平分∠ABC,BE⊥AC,CF⊥A∴OE＝OF（角平分线性质）,∠BFC＝∠CEB＝90∵∠BOF＝∠COE∴△BOF≌△COE（ASA）∴BF＝CE或∵AD平分∠ABC∴∠BAO＝∠CA

连接PC,∵AB=AC,AD⊥BC,∴AD是BC的垂直平分线,PB=PC；还有∠PBC=∠PCB；∵AB=AC,∴∠ABC=∠ACB；还有∠ABP=∠ACP；∵CF∥AB,∴∠ABP=∠PFC=∠AC

∵AE=CF,∴AE+EF=EF+FC,即AF=CE,∵DE⊥AC,BF⊥AC,∴∠AFB=∠CED=90°,又∵AB=CD,∴RT△ABF≌RT△CDE（HL）,∴∠A=∠C,∴AB∥CD.

证明：∵BF⊥AC,CE⊥AB∴∠AEC＝∠AFB＝90,∠BFC＝∠CEB＝90∵BE＝CF,∠BDE＝∠CDF∴△BDE≌△CDF（AAS）∴DE＝DF∵AD＝AD∴△ADE≌△ADF（HL）∴∠

证明：∵BF⊥AC，CE⊥AB，∴∠BED=∠CFD=90°．在△BED和△CFD中，∠BED＝∠CFD∠BDE＝∠CDFBE＝CF，∴△BED≌△CFD（AAS），∴DE=DF．∵DF⊥AC，DE⊥