问题补充:
已知:如图,在等边三角形ABC中,D、E分别为BC、AC上的点,且AE=CD,连接AD、BE交于点P,作BQ⊥AD,垂足为Q.求证:BP=2PQ.
答案:
证明:∵△ABC是等边三角形,
∴AB=AC=BC,∠C=∠ABC=60°,
∵AE=CD,
∴EC=BD;
∴△BEC≌△ADB(SAS),
∴∠EBC=∠BAD;
∵∠ABE+∠EBC=60°,则∠ABE+∠BAD=60°,
∵∠BPQ是△ABP外角,
∴∠ABP+∠BAP=60°=∠BPQ,
又∵BQ⊥AD,
∴∠PBQ=30°,
∴BP=2PQ.
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