HMMT 二月 2008 · TEAM1 赛 · 第 11 题

英文解析

11. [ 30 ] Let lines BI and EF meet at K . Show that I, K, E, C, D are concyclic.
    Solution: First, note that there are two possible configurations, as K could lie insidesegment EF , or on its extension. The following proof works for both cases. We have
    I EKFA
    B C
    4D
    1 1 1
    °
    ∠ KIC = ∠ IBC + ∠ ICB = ∠ ABC + ∠ ACB = 90 − ∠ BAC = ∠ AEF.
    2 2 2
    It follows that I, K, E, C are concyclic. The point D also lies on this circle because ∠ IDC =
    °
    ∠ IEC = 90 . Thus, all five points are concyclic.