一道数学几何体,答案快点来

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  • 可以用同一法结合面积证明.

    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.