如图,在△ABC中,已知AB=BC=CA,AE=CD,AD与BE交于点P,BQ⊥AD于点Q,求证:BP=2PQ.

问题描述:

如图,在△ABC中,已知AB=BC=CA,AE=CD,AD与BE交于点P,BQ⊥AD于点Q,求证:BP=2PQ.

证明:∵AB=BC=CA,∴△ABC为等边三角形,∴∠BAC=∠C=60°,在△ABE和△CAD中AB=AC∠BAC=∠CAE=DC∴△ABE≌△CAD(SAS),∴∠ABE=∠CAD,∵∠BPQ=∠ABE+∠BAP,∴∠BPQ=∠CAD+∠BAP=∠CAB=60°,∵BQ⊥AD∴∠BQ...