本文将给出三角形等角共轭点的一个新性质,即命题　设P、Q是△ABC的等角共轭点(∠PAB=∠QAC,∠PBC=∠QBA,∠PCB=∠QCA),则有AP.AQAB.AC+BP.BQBA.BC+CP.CQCA.CB=1.证明　如图1,设D是射线AQ上的点,且使得满足∠ACD=∠APB.因为∠APB∠ACB,则点D必在△ABC的外部.又因∠PAB=∠CAD, In this paper, we will give a new property of the triangle iso-angle conjugate point, that is, the proposition that P and Q are △ABC equidistant conjugate points (∠PAB=∠QAC,∠PBC=∠QBA,∠PCB=∠QCA), Then there is AP.AQAB.AC+BP.BQBA.BC+CP.CQCA.CB=1. Prove that as shown in FIG. 1, let D be the point on the ray AQ, and let ∠ACD=∠APB. Because ∠APB∠ACB , point D must be outside the △ABC. Also because of ∠PAB=∠CAD,