已知在△ABC中,∠BAC=90º,AB=AC,P为BC上任意一点,求证:BP²+CP²=2AP²
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![已知在△ABC中,∠BAC=90º,AB=AC,P为BC上任意一点,求证:BP²+CP²=2AP²](/uploads/image/z/11330995-67-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0BAC%3D90%26%23186%3B%2CAB%3DAC%2CP%E4%B8%BABC%E4%B8%8A%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9ABP%26%23178%3B%2BCP%26%23178%3B%3D2AP%26%23178%3B)
已知在△ABC中,∠BAC=90º,AB=AC,P为BC上任意一点,求证:BP²+CP²=2AP²
已知在△ABC中,∠BAC=90º,AB=AC,P为BC上任意一点,求证:BP²+CP²=2AP²
已知在△ABC中,∠BAC=90º,AB=AC,P为BC上任意一点,求证:BP²+CP²=2AP²
过P点分别做PM、PN垂直于AB、AC,垂足分别为M、N
四边形AMPN为矩形
MB=MP=AN NC=NP=AM
BP² = 2MP² CP² = 2NP²
因为 AM²+MP² = AP²
NP²+MP² = AP²
所以 BP²+CP² =2(MP²+NP²)=2AP²