文[1]给出了周界中点三角形的边与旁切圆半径的一个性质,将其延拓至三角形的外角平分线三角形中,得到图1定理如图1,△DEF是△ABC的三条外角平分线三角形.设BC=a,CA=b,AB=c,△BCD,△ACE,△ABF的内切圆半径分别为rA,rB,rC,则a rA+brB+crC≥63.证明由文[2]知,a=rA.cosA2+cosB2+cos In [1], we give a property of the radius of the edge and the tangent circle of the midpoint triangle of the perimeter, and extend it to the triangle of the external bisector of the triangle. The theorem of Figure 1 is shown in Figure 1. △DEF is the three outer angles of △ABC. The bisector triangle. Let BC = a, CA = b, AB = c, △ BCD, △ ACE, △ ABF inscribed circle radius respectively rA, rB, rC, then a rA + brB + crC ≥ 63. proved by [2] knows that a=rA.cosA2+cosB2+cos