Question

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

Answer 1

设AB中点M,CD中点N

PA*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+PB2+PC2+PD2 = 2PM^2 + 2PN^2 = 2PM^2 + 2OM^2 = 2R^2

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