Question

如图，三角形ABC，分别在AB，AC上取点D，E 使得BD=CE,连接BE,CD相交于点P，点M是BC的中点，∠BAC…请看图

Answer 1

在射线PM上取Q', 使PM = MQ', 连AQ', BQ', CQ', DQ', EQ'.

∵BM = MC, PM = MQ',

∴BPCQ'是平行四边形, 即有CP // BQ', BP // CQ',

∴SΔDBQ' = SΔCBQ' = SΔCEQ'.

又∵BD = CE,

∴Q'到AB的距离 = 2·SΔDBQ'/BD = 2·SΔCEQ'/CE = Q'到AC的距离,

∴Q'在∠BAC的平分线AQ上.

于是Q'为PM与AQ的交点, 即Q'与Q重合.

故BPCQ即BPCQ', 已证为平行四边形.

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http://zhidao.baidu.com/question/577499994