如图,ΔABC中,AB=AC,∠BAC=90º,D是AB中点,连CD,AE⊥CD于F交BC于E点。求证:∠BDE=∠EAC
如图,ΔABC中,AB=AC,∠BAC=90º,D是AB中点,连CD,AE⊥CD于F交BC于E点。求证:∠BDE=∠EAC
过B作BG垂直于AB交AE延长线于G,易证△ABG≌△CAD,∴∠ADC=∠G BG=AD=BD,∠DBC=∠GBE=45°,BE=BE, ∴△EBD≌△EBG,,∴∠BDE=∠G, 又,∠CAF+DAF=∠DAF+∠ADF=90°, ∴∠CAF=∠ADC, ∴∠BDE=∠EAC
答:证明:(1)如图作AG平分∠BAC,交BD于点G ,交BC于H AE交BD于F ∵∠BAC=90°,AE⊥BD ∴∠DAF+∠ADB=∠ABD+∠ADB=90°...详情>>
答:I came.详情>>