2015/G1 圆 ω 1 \omega_1 ω1 与圆 ω 2 \omega_2 ω2 内切于点 T T T. M M M, N N N 是圆 ω 1 \omega_1 ω1 上不同于 T T T 的不同两点. 圆 ω 2 \omega_2 ω2 的两条弦 A B AB AB, C D CD CD 分别过 M M M, N N N. 证明: 若线段 A C AC AC, B D BD BD, M N MN MN 交于同一点 K K K. 求证: T K T K TK 平分 ∠ M T N \angle MTN ∠MTN.
证明: 只需证 M K / N K = M T / N T MK/NK=MT/NT MK/NK=MT/NT.
T M 2 / M A ⋅ M B = T N 2 / B D ⋅ N C TM^2/MA \cdot MB=TN^2/BD \cdot NC TM2/MA⋅MB=TN2/BD⋅NC.
M A ⋅ M B / M K 2 = sin ∠ M K A sin ∠ B A K sin ∠ M K B sin ∠ K B A MA \cdot MB/MK^2=\frac{\sin \angle MKA}{\sin \angle BAK}\frac{\sin\angle MKB}{\sin \angle KBA} MA⋅MB/MK2=sin∠BAKsin∠MKAsin∠KBAsin∠MKB.
类似地, 有 N D ⋅ N C / N K 2 = sin ∠ N K D sin ∠ K C D sin ∠ N K C sin ∠ K C D ND\cdot NC/NK^2=\frac{\sin \angle NKD}{\sin \angle KCD}\frac{\sin \angle NKC}{\sin \angle KCD} ND⋅NC/NK2=sin∠KCDsin∠NKDsin∠KCDsin∠NKC.
∠ K B A = ∠ K D C \angle KBA=\angle KDC ∠KBA=∠KDC, ∠ K B A = ∠ K C D \angle KBA=\angle KCD ∠KBA=∠KCD.
∠ M K B = ∠ N K D \angle MKB=\angle NKD ∠MKB=∠NKD, ∠ M K A = ∠ K N C \angle MKA=\angle KNC ∠MKA=∠KNC.
所以 M A ⋅ M B / M K 2 = N D ⋅ N C / N K 2 MA \cdot MB/MK^2=ND\cdot NC/NK^2 MA⋅MB/MK2=ND⋅NC/NK2, 进而 M K 2 / N K 2 = M T 2 / N T 2 MK^2/NK^2=MT^2/NT^2 MK2/NK2=MT2/NT2, 即 M K / N K = M T / N T MK/NK=MT/NT MK/NK=MT/NT.
证毕.
