Question

如图,三角形ABC为等边三角形,D为三角形内一点,且有DA=DB,BP=BA,角BPD=30度,求证BD平分角PBC

Answer 1

连接CD、CP.因为,DA = DB ,CA = CB ,所以,D和C都在AB的垂直平分线上,可得：∠BCD = (1/2)∠BCA = 30° = ∠BPD .因为,BP = BA = BC ,所以,∠BCP = ∠BPC ；可得：∠DCP = ∠BCP-∠BCD = ∠BPC-∠BPD = ∠DPC ,所以,D...