先證一個引理: 圓外切四邊形對角線以及對邊切點連線四線共點.如下圖, 設ABCD是圓外切四邊形, K, L, M, N是其內切圓與各邊的切點,KL交BD於G, MN交BD於G", 以下證明G與G"重合.延長ABCD對邊, 得到兩個交點E, F.∵K, G, L共線,∴BG/GD·DK/KE·EL/LB = 1 (對△BDE使用Menelaus定理).而∵KE = EL (切線長定理),∴BG/GD = LB/DK.另一方面, ∵M, G", N共線,∴BG"/G"D·DN/NF·FM/MB = 1 (對△BDF使用Menelaus定理).而∵NF = FM (切線長定理),∴BG"/G"D = MB/DN.又∵LB = MB, DK = DN (切線長定理),∴BG/GD = BG"/G"D,故G與G"重合, 也即BD經過KL與MN的交點.同理AC也經過KL與MN的交點, 因此BD, AC, KL, MN四線共點.如下圖, 考慮原題圖形的一個區域性.設EL交AB於G, 內切圓與AB, AC分別相切於P, Q.連GC, PQ, BE, 由引理知三線共點, 設為點O.記BC = a, CA = b, AB = c.由切線長定理不難算得AP = QA = (b+c-a)/2,進而PB = (c+a-b)/2, QE = QA-EA = (c-a)/2.∵P, O, Q共線,∴BO/OE·EQ/QA·AP/PB = 1 (對△BEA使用Menelaus定理),∴BO/OE = PB/EQ = (c+a-b)/(c-a).又∵G, O, C共線,∴BO/OE·EC/CA·AG/GB = 1 (對△BEA使用Menelaus定理),∴GB/AG = BO/OE·1/2 = (c+a-b)/(2c-2a),∴GF/AF = (AF-AG)/AF = (AB-2AG)/AB = ((AG+GB)-2AG)/(AG+GB)= (GB-AG)/(AG+GB) = ((c+a-b)-(2c-2a))/((2c-2a)+(c+a-b)) = (3a-b-c)/(3c-a-b).而∵DE // AB, DL // AC (中位線定理),∴FL/LD = GL/LE = GF/FA = (3a-b-c)/(3c-a-b).同理, 在原題圖形中可得DM/ME = (3b-c-a)/(3a-b-c), EI/IF = (3c-a-b)/(3b-c-a).故FL/LD·DM/ME·EI/IF = 1.對△FDE使用Menelaus定理即得I, L, M共線, 證畢.
先證一個引理: 圓外切四邊形對角線以及對邊切點連線四線共點.如下圖, 設ABCD是圓外切四邊形, K, L, M, N是其內切圓與各邊的切點,KL交BD於G, MN交BD於G", 以下證明G與G"重合.延長ABCD對邊, 得到兩個交點E, F.∵K, G, L共線,∴BG/GD·DK/KE·EL/LB = 1 (對△BDE使用Menelaus定理).而∵KE = EL (切線長定理),∴BG/GD = LB/DK.另一方面, ∵M, G", N共線,∴BG"/G"D·DN/NF·FM/MB = 1 (對△BDF使用Menelaus定理).而∵NF = FM (切線長定理),∴BG"/G"D = MB/DN.又∵LB = MB, DK = DN (切線長定理),∴BG/GD = BG"/G"D,故G與G"重合, 也即BD經過KL與MN的交點.同理AC也經過KL與MN的交點, 因此BD, AC, KL, MN四線共點.如下圖, 考慮原題圖形的一個區域性.設EL交AB於G, 內切圓與AB, AC分別相切於P, Q.連GC, PQ, BE, 由引理知三線共點, 設為點O.記BC = a, CA = b, AB = c.由切線長定理不難算得AP = QA = (b+c-a)/2,進而PB = (c+a-b)/2, QE = QA-EA = (c-a)/2.∵P, O, Q共線,∴BO/OE·EQ/QA·AP/PB = 1 (對△BEA使用Menelaus定理),∴BO/OE = PB/EQ = (c+a-b)/(c-a).又∵G, O, C共線,∴BO/OE·EC/CA·AG/GB = 1 (對△BEA使用Menelaus定理),∴GB/AG = BO/OE·1/2 = (c+a-b)/(2c-2a),∴GF/AF = (AF-AG)/AF = (AB-2AG)/AB = ((AG+GB)-2AG)/(AG+GB)= (GB-AG)/(AG+GB) = ((c+a-b)-(2c-2a))/((2c-2a)+(c+a-b)) = (3a-b-c)/(3c-a-b).而∵DE // AB, DL // AC (中位線定理),∴FL/LD = GL/LE = GF/FA = (3a-b-c)/(3c-a-b).同理, 在原題圖形中可得DM/ME = (3b-c-a)/(3a-b-c), EI/IF = (3c-a-b)/(3b-c-a).故FL/LD·DM/ME·EI/IF = 1.對△FDE使用Menelaus定理即得I, L, M共線, 證畢.