Question

如图,○O半径为根号17,AB为直径,M、N为圆上两点,M、N在AB同侧,AM、BN交于P,则AP·AM+BP·BN=?

Answer 1

分析：连接AN、BM,根据圆周角定理,由AB是直径,可证∠AMB=90°,由勾股定理知,BP 2 =MP 2 +BM 2 ,由相交弦定理知,AP•PM=BP•PN,原式=AP（AP+PM）+BP（BP+PN）=AP 2 +AP•PM+BP 2 +BP•PN=AP 2 +BP 2 +2AP•PM=AP 2 +MP 2 +BM 2 +2AP•PM=AP 2 +（AP+PM） 2 =AP 2 +AM 2 =AB 2 =36．

连接AN、BM,

∵AB是直径,

∴∠AMB=90°．

∴BP^2 =MP^2 +BM^2

∴AP•PM=BP•PN

原式=AP（AP+PM）+BP（BP+PN）=AP 2 +AP•PM+BP^2 +BP•PN

=AP^2 +BP^2 +2AP•PM

=AP^2 +MP^2 +BM^2 +2AP•PM

=BM^2 +（AP+PM）^2 =BM^2+AM^2 =AB^2 =4*17=68．