Question

如图,在△ABC中,∠B＝2∠C,AD⊥BC于点D,AB＝6,AP＝BP＝3,PM平行于AC交BC于点M,则DM的长为多

Answer 1

AB=6,PA=PB=3,则点P为AB的中点.

连接PD.AD⊥BC,

则:PD=PB=3,∠PDB=∠B=2∠C;

PM∥AC,则∠PMD=∠C,故∠PDB=2∠PMD.

又∵∠PDB=∠DPM+∠PMD.

∴∠DPM+∠PMD=2∠PMD.

故:∠DPM=∠PMD,DM=PD=3.