证明:∵∠BAC=90°,∴∠CAF=90°=∠BAC,∴∠FCA+∠F=90°,∵CF⊥BD交BD的延长线于点E,∴∠BEF=90°,∴∠ABD+∠F=90°,∴∠ABD=∠ACF.在△ABD与△ACF中,∵∠BAC=∠CAF,AB=AC,∠ABD=∠ACF,∴△ABD≌△ACF(ASA),∴BD=CF.
∵∠bac=90∴∠abd+∠adb=90,∠caf=∠bac=90∵∠cde=∠adb∴∠abd+∠cde=90∵cf⊥bd∴∠acf+∠cde=90∴∠abd=∠acf∵ab=ac∴△abd≌△acf (asa)∴bd=cf
证明:∵∠BAC=90∴∠ABD+∠ADB=90,∠CAF=∠BAC=90∵∠CDE=∠ADB∴∠ABD+∠CDE=90∵CF⊥BD∴∠ACF+∠CDE=90∴∠ABD=∠ACF∵AB=AC∴△ABD≌△ACF (ASA)∴BD=CF
利用相似三角形∵∠BAD=∠CED=RT∠,∠BDA=∠CDE∴∠ABD=∠DCE又∵∠BAD=∠CAF=RT∠,AB=AC∴△ABD≌△ACF∴BD=CF求采纳!
延长CE交AB于点F可证△ABD≌△ACF∴BD=CF∵BD是角平分线,∠BEC=90°∴△BFE≌△BCE EF=CE又∵CF=2CE∴BD=2CE
∵∠BAC的平分线是BD, CE⊥BE. ∴∠MBE=∠EBC, ∠BEM=∠BEC=90° ∵BE=BE ∴△BEM≌△BEC(AAS) ∴EM=CE 即CM=2CE…………………………① ∵∠ABE的平分线是BE, AB=AC, ∠BAC=90° ∴∠ABD=DBC=22.5°, ∠BCA=45° ∵CE⊥BE ∴∠EBC+∠BCD+∠DCE=90° ∴∠DCE=22.5° ∴∠ABD=∠DCE 又∵AB=AC,且∠BAC=∠MAC=90° ∴△ABD≌△ACM (ASA) ∴BD=CM………………………………② 由①②得,BD=2CE.
证明:分别延长BA、CE相交于O,∵BE⊥CE,∴∠BEC=∠BEO=90°,∵BD平分∠ABC,∴∠EBC=∠EBO,∵BE=BE,∴ΔBEC≌ΔBEO,∴CE=OE,∴CO=2CE,∵BE⊥OC,∴∠ABD+∠O=90°,∵∠BAC=90°,∴∠ACO+∠O=90°,∴∠ABD=∠ACO,在RTΔABD与RTΔACO中,∠BAD=∠CAO=90°,AB=AC,∠ABD=∠ACO,∴ΔABD≌ΔACO,∴BD=CO=2CE.
证明:延长BA、CE交于F∵∠BAC=90°,BE⊥CF,∠ADB=∠CDE∴∠CAF=90°,∠ABD=∠ACF∵在△BAD和△CAF中, ∠ABD=∠ACF,AB=AC,∠BAC=∠CAF∴△BAD≌△CAF(ASA)∴BD=CF∵BE是∠ABC的平分线,BE⊥CF∴CE=EF,即CF=2CE∴BD=2CE
从c作cm⊥ac,交af延长线于m.连结bm, 在△amc和△adb中, ∵△abc是等腰rt△. ∴ab=ac, ∵<aed=90°, ∴<dae+<ade=90°, ∵<bad=90°, ∴<abd+<adb=90, ∴<abd=<dae=<cam, ∴△abd≌△cam, ∴ad=cm,<adb=<amc, ∵ad=cd(已知), ∴dc=cm, 在△cdf和△cmf中, ∵<fcm=90-<dcf=90°-45°=45°, <fcm=<dcf, fc=fc(公用边), dc=cm, ∴△dcf≌△mcf, ∴<fmc=<fdc, ∴<adb=<cdf. 证毕.
题目是不是这样:“已知,三角形ABC中,∠BAC=90°.AB=AC,BD是ABC的角平分线,∠1是∠ABD,∠2是∠CBD,CE垂直BD交BD延长线于E,求证:BD=2CE” 如果是的话,证明如下:延长BA、CE,两线相交于点F∵CE⊥BD∴∠BEF=∠BEC=90°在△BEF和△BEC中∠1=∠2, BE=BE, ∠BEF=∠BEC∴△ABD≌△ACF(ASA)∴EF=EC∴CF=2CE∵∠1+∠ADB=90°,∠ACE+∠CDE=90°又∵∠ADB=∠CDE∴∠1=∠ACE在△ABD和△ACF中∠1=∠ACE, AB=AC, ∠BAD=∠CAF=90°∴△ABD≌△ACF(ASA)∴BD=CF∴BD=2CE