△ A B C \triangle ABC △ABC 的外心为点 O O O, 外接圆为 Γ \Gamma Γ. 射线 A O AO AO, B O BO BO, C O CO CO 分别交 Γ \Gamma Γ 于点 D D D, E E E, F F F. X X X 是 △ A B C \triangle ABC △ABC 内部的一点. 射线 A X AX AX, B X BX BX, C X CX CX 分别交 Γ \Gamma Γ 于点 A 1 A_1 A1, B 1 B_1 B1, C 1 C_1 C1. 射线 D X DX DX, E X EX EX, F X FX FX 分别与 Γ \Gamma Γ 交于 D 1 D_1 D1, E 1 E_1 E1, F 1 F_1 F1. 求证: 三条直线 A 1 D 1 A_1D_1 A1D1, B 1 E 1 B_1E_1 B1E1, C 1 F 1 C_1F_1 C1F1 三线共点. (《高中数学联赛模拟试题精选》"学数学"系列第4套)
证实: 由角元形式的塞瓦定理, 只需证实:
sin ∠ C 1 F 1 D 1 sin ∠ C 1 F 1 E 1 sin ∠ A 1 D 1 E 1 sin ∠ A 1 D 1 F 1 sin ∠ B 1 E 1 F 1 sin ∠ B 1 E 1 D 1 = 1 \frac{\sin \angle C_1F_1D_1}{\sin \angle C_1F_1E_1}\frac{\sin \angle A_1D_1E_1}{\sin \angle A_1D_1F_1}\frac{\sin \angle B_1E_1F_1}{\sin \angle B_1E_1D_1}=1 sin∠C1F1E1sin∠C1F1D1sin∠A1D1F1sin∠A1D1E1sin∠B1E1D1sin∠B1E1F1=1.
此中:
sin ∠ C 1 F 1 D 1 sin ∠ C 1 F 1 E 1 = C 1 D 1 C 1 E 1 \frac{\sin \angle C_1F_1D_1}{\sin \angle C_1F_1E_1}=\frac{C_1D_1}{C_1E_1} sin∠C1F1E1sin∠C1F1D1=C1E1C1D1, sin ∠ A 1 D 1 E 1 sin ∠ A 1 D 1 F 1 = E 1 A 1 A 1 F 1 \frac{\sin \angle A_1D_1E_1}{\sin \angle A_1D_1F_1}=\frac{E_1A_1}{A_1F_1} sin∠A1D1F1sin∠A1D1E1=A1F1E1A1, sin ∠ B 1 E 1 F 1 sin ∠ B 1 E 1 D 1 = F 1 B 1 B 1 D 1 \frac{\sin \angle B_1E_1F_1}{\sin \angle B_1E_1D_1}=\frac{F_1B_1}{B_1D_1} sin∠B1E1D1sin∠B1E1F1=B1D1F1B1.
所以这等价于:
C 1 D 1 C 1 E 1 E 1 A 1 A 1 F 1 F 1 B 1 B 1 D 1 = 1 \frac{C_1D_1}{C_1E_1}\frac{E_1A_1}{A_1F_1}\frac{F_1B_1}{B_1D_1}=1 C1E1C1D1A1F1E1A1B1D1F1B1=1.
C 1 D 1 C 1 E 1 = C D D 1 X C X C E E 1 X C X = C D C E D 1 X E 1 X \frac{C_1D_1}{C_1E_1}=\frac{CD \frac{D_1X}{CX}}{CE\frac{E_1X}{CX}}=\frac{CD}{CE}\frac{D_1X}{E_1X} C1E1C1D1=CECXE1XCDCXD1X=CECDE1XD1X.
类似地, 可求出:
E 1 A 1 A 1 F 1 = A E A F E 1 X F 1 X \frac{E_1A_1}{A_1F_1}=\frac{AE}{AF}\frac{E_1X}{F_1X} A1F1E1A1=AFAEF1XE1X.
F 1 B 1 B 1 D 1 = B F B D F 1 X D 1 X \frac{F_1B_1}{B_1D_1}=\frac{BF}{BD}\frac{F_1X}{D_1X} B1D1F1B1=BDBFD1XF1X.
三式相乘, 得:
C 1 D 1 C 1 E 1 E 1 A 1 A 1 F 1 F 1 B 1 B 1 D 1 = C D C E A E A F B F B D = C D B D B F A F A E C E \frac{C_1D_1}{C_1E_1}\frac{E_1A_1}{A_1F_1}\frac{F_1B_1}{B_1D_1}=\frac{CD}{CE}\frac{AE}{AF}\frac{BF}{BD}=\frac{CD}{BD}\frac{BF}{AF}\frac{AE}{CE} C1E1C1D1A1F1E1A1B1D1F1B1=CECDAFAEBDBF=BDCDAFBFCEAE.
由角元形式的塞瓦定理可知:
sin ∠ C A D sin ∠ B A D sin ∠ B C F sin ∠ A C F sin ∠ A B E sin ∠ C B E = 1 \frac{\sin \angle CAD}{\sin \angle BAD}\frac{\sin \angle BCF}{\sin \angle ACF}\frac{\sin \angle ABE}{\sin \angle CBE}=1 sin∠BADsin∠CADsin∠ACFsin∠BCFsin∠CBEsin∠ABE=1
此中: sin ∠ C A D sin ∠ B A D = C D B D \frac{\sin \angle CAD}{\sin \angle BAD}=\frac{CD}{BD} sin∠BADsin∠CAD=BDCD, sin ∠ B C F sin ∠ A C F = B F A F \frac{\sin \angle BCF}{\sin \angle ACF}=\frac{BF}{AF} sin∠ACFsin∠BCF=AFBF, sin ∠ A B E sin ∠ C B E = A E C E \frac{\sin \angle ABE}{\sin \angle CBE}=\frac{AE}{CE} sin∠CBEsin∠ABE=CEAE.
所以 C D B D B F A F A E C E = 1 \frac{CD}{BD}\frac{BF}{AF}\frac{AE}{CE}=1 BDCDAFBFCEAE=1.
证毕.
完稿时间: 2025年5月3日.
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发表于 2025-5-4 05:49:55
