在半径为R的圆O中,弦AB垂直于弦CD于点P.求证：AP∧2+CP∧2+PB∧2+PD^2为定值
问题描述：
在半径为R的圆O中,弦AB垂直于弦CD于点P.求证：AP∧2+CP∧2+PB∧2+PD^2为定值
答
设AB中点M,CD中点NPA*PB = AM^2 - PM^2 = AO^2 - PO^2 = R^2 - PO^2同理：PC*PD = R^2 - PO^2则PA^2+PB^2 = AB^2 - 2PA*PB = 2(R^2 - OM^2) - 2(R^2 - PO^2) = 2(PO^2 - OM^2) = 2PM^2同理PC^2+PD^2 = 2PN^2=>PA2+PB...太快了!学神～为什么PA*PB=AM^2-PM^2????