一道数学几何体,答案快点来
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/13 06:03:41
一道数学几何体,答案快点来
![一道数学几何体,答案快点来](/uploads/image/z/18155245-13-5.jpg?t=%E4%B8%80%E9%81%93%E6%95%B0%E5%AD%A6%E5%87%A0%E4%BD%95%E4%BD%93%2C%E7%AD%94%E6%A1%88%E5%BF%AB%E7%82%B9%E6%9D%A5)
可以用同一法结合面积证明.
![](http://img.wesiedu.com/upload/5/53/5532ad4333937e0302b0971171de0408.jpg)
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.
![](http://img.wesiedu.com/upload/5/53/5532ad4333937e0302b0971171de0408.jpg)
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.