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.