问题补充:
如图,△ABC中,AD⊥BC于D,BE⊥AC于E,AD交BE于F,BF=AC,则∠ABC等于A.40°B.45°C.60°D.30°
答案:
B解析∵AD⊥BC于D,BE⊥AC于E,∴∠EAF+∠AFE=90°,∠DBF+∠BFD=90,又∵∠BFD=∠AFE(对顶角相等),∴∠EAF=∠DBF,在Rt△ADC和Rt△BDF中, ∠EAF=∠DBF, ∠FDB=∠CDA, AC=BF∴△ADC≌△BDF,∴BD=AD,即∠ABC=∠BAD=45°.故选B.
时间:2019-08-25 17:36:26
如图,△ABC中,AD⊥BC于D,BE⊥AC于E,AD交BE于F,BF=AC,则∠ABC等于A.40°B.45°C.60°D.30°
B解析∵AD⊥BC于D,BE⊥AC于E,∴∠EAF+∠AFE=90°,∠DBF+∠BFD=90,又∵∠BFD=∠AFE(对顶角相等),∴∠EAF=∠DBF,在Rt△ADC和Rt△BDF中, ∠EAF=∠DBF, ∠FDB=∠CDA, AC=BF∴△ADC≌△BDF,∴BD=AD,即∠ABC=∠BAD=45°.故选B.
已知AC=FE BC=DE 点A D B F在一条直线上 AD=BF.求证:∠E=∠C.
2023-11-04