题目一个圆通过△ABC的顶点A分别交线段AC、BC于点D、E,直线B ED交于点F,直线BD和CF交于点M明:MF=MC的充要条件为MB·MD=MC2[1].(2003,美国数学奥林匹克)本文给出一种较简捷的证法.图1证明:如图1,联结AE.在△CBF中,因为AC、BM、EF交于点D,由塞瓦定理可得CEBE·ABAF·MFMC=1.① A circle is intersected by a vertex A of △ABC and a line AC and BC are at points D and E. The line B ED intersects at point F. The line BD and CF intersect at point M. The necessary and sufficient condition for MF=MC is MB·MD. =MC2[1].(2003, American Mathematical Olympiad) This paper presents a simpler proof method. Figure 1 proves that: As shown in Figure 1, linking AE. In △CBF, AC, BM, and EF are intersected at point D. , available from CEVA’s theorem CEBE·ABAF·MFMC=1.1