《数学教学》2014年第10期《数学问题与解答栏目》刊登的第926号问题是:已知点I为△ABC的与BC边相切的旁切圆圆心,求证:(IA~2)/(CA·AB)-(IB~2)/(AB·BC)-(IC~2)/(BC·CA)=1.经查阅资料,发现该问题也是文[1]命题1对应的结论,并且文[1]最终把它推广到椭圆:设P、Q是△ABC的一个分别与BC边,AC、AB延长线相切的旁切椭圆的两个焦点,则下列等式恒成立: Problem 926 published in the column of “Mathematical Problems and Solutions”, Volume 10, 2014 is that the point I is known as △ ABC and the tangent circular center tangent to BC, so that: (IA ~ 2) / (CA · AB) - (IB ~ 2) / (AB · BC) - (IC ~ 2) / (BC · CA) = 1. After reviewing the data, we find that this problem is also the conclusion of Proposition 1 in [1] , And the text [1] eventually extends it to the ellipse: Let P and Q be the two focal points of a parsimonious ellipse tangent to the ABC, AC, and AB extension, respectively, then the following equation holds: