一、在锐角△ABC中 ,AD是∠A的内角平分线 ,点D在边BC上 ,过点D分别作DE⊥AC、DF⊥AB ,垂足分别为E、F ,连结BE、CF ,它们相交于点H ,△AFH的外接圆交BE于点G .求证 :以线段BG、GE、BF组成的三角形是直角三角形 .二、设A {0 ,1,2 ,… ,2 9},满足 :对任何整数k及A中任意数a、b(a First, in the acute angle △ABC, AD is the internal angle bisector of ∠A, point D is on the side BC, and the point D is DE ⊥ AC, DF ⊥ AB , respectively, and the feet are E and F respectively, which link BE and CF. They intersect at point H. The circumcircle of △AFH intersects BE at point G. Proof: The triangle formed by line segment BG, GE, and BF is a right-angled triangle. Second, let A {0, 1, 2, ......, 2 9}, Satisfaction: For any integer k and A in any number a, b (a