Question

已知在等边三角形ABC中，D，E分别是BC，AC上的点，且AE=CD，连接AD、BE交与点P，做BQ⊥AD，垂足为Q。求证

已知在等边三角形ABC中，D，E分别是BC，AC上的点，且AE=CD，连接AD、BE交与点P，做BQ⊥AD，垂足为Q。求证BP＝2PQ

Answer 1

因为△BEA≌△ADC. ∠APE=∠ABE+∠BAP ∠ABE=CAD 由于是等边三角形∠BAD+∠CAD=60° 所以∠BAD+∠ABD=60° 所以∠APE＝60° =∠BPQ 剩下的你自己该知道怎么回事了吧。直角三角形有一个角为60。

Answer 2

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