文[1]指出:设△ABC的三个内角A、B、C的对边分别为a、b、c,内心为O,内切圆半径为r,M为△ABC内切圆上任意一点,ME⊥AB于E,MF⊥AC于F,则2r-r(a-b+c)bc(a+b-c)≤ME+MF≤2r+r(a-b+c)b(ca+b-c).①笔者经思考发现,在双圆n边形中有定理设双圆n边形A1A2…An的内心为I,内切圆半径为r,P为 The article [1] pointed out that the opposite sides of the three inner angles A, B, and C of △ABC are a, b, and c respectively, the inner center is O, the radius of the inscribed circle is r, and M is △ABC, any point on the inscribed circle. , ME ⊥ AB in E, MF ⊥ AC in F, then 2r-r (a-b + c) bc (a + bc) ≤ ME + MF ≤ 2r + r (a-b + c) b (ca + bc ).1 The author found out that in the double-circular n-sided form, the theorem sets the inner circle of the double-circular n-sided A1A2...An to I, and the radius of the inscribed circle is r, P is