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

解: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.