Question

在△ABC中,AD⊥BC,BE⊥AC,AD、BE相交于F,连接CE且AC=BF.求证：∠ABC+∠FCB=90°

Answer 1

证明:∠BDF=∠AEF=90°;∠BFD=∠AFE;则:∠FBD=∠FAE.

又∠BDF=∠ADC=90°;是BF=AC.

故:⊿BDF≌ΔADC(ASA).

所以:BD=AD,∠ABC=∠BAD=45°;

同理:DF=DC,∠DFC=∠FCB=45°.

∠ABC+∠FCB=90°.