Question

如图,已知B（-1,0）,C（1,0）,A为y轴正半轴上一点,点D为第二象限一动点,E在BD的延长线上,CD交AB

Answer 1

∵∠BDC=∠FAC,∠BFD=∠CFA,∴∠DBF=∠ACF,△BDF∽△CAF,DF;CF=BF;AF,又∠DFC=∠BFC,∴△DFA∽△BFC,∴∠ADF=∠CBF,又∠EDC=∠EBA+∠ABC+∠BCD=∠ACF+∠ABC+∠BCD=∠ABC+∠ACB,∴∠EDA=∠ACB,又BO=CO,AO⊥BC,∴∠ABC=∠ACB,∴∠EDA=∠CDA,即AD平分∠CDE