等边三角形ABC的边AC上有一动点D,延长AB至E,使BE=CD,连接DE,交BC于点P.求证：DP=PE

作DG//AE∴∠ABC=∠DGC ∠E=∠EDG∵ 等边△ABC∴ ∠ABC=∠C∵ ∠ABC=∠DGC ∠ABC=∠C∴ ∠DGC=∠C ∴DG=CD∵ BE=CD ∴ DG=BE在△BEP和△GDP中∠E=∠EDG∠BPE=∠DPGBE=DG...