证明：∵AB＝BC＝AC∴等边△ABC∴∠BAC＝∠ABC＝∠ACB＝60∵AE＝CD∴△ABE≌△CAD （SAS）∴∠ABE＝∠CAD∴∠BPD＝∠ABE+∠BAD＝∠CAD+∠BAD＝∠BAC＝60∵BQ⊥AD∴BP＝2PQ
