Thứ Sáu, 12 tháng 9, 2014

32- Gacia Capitan's line-9 points are collinear

Today I recheck a problems of me at here: ADPGEP#624 , I generalization following:

Let ABC be a triangle, a line (d) on the plane, (d) meets three sides BC,CA,AB respectively at A_1,B_1,C_1. Let P is any point on the plane. PA_1meets AC,AB respectively at A_b,A_c. Define B_c,B_a,C_a,C_b cyclically.

A_bB_a meets AB at C_2, define B_2,A_2 cyclically.

Let P_a be a points on the line PA_1, such that \frac{P_aA_b}{P_aA_c}=-\frac{PA_b}{PA_c}; define P_b,P_ccyclically.



I named the line through P_a,P_b,P_c is Gacia Capitan's line

Please see also: Garcia Capitan blog

Now we have A_2,B_2,C_2 also lie on Gacia Capitan's line.

- Further more: Two triangle A_bB_cC_a and A_cB_aC_b are perpective, perpector is P(please see the figure attachment). A_bB_c meets A_cB_a at MB_cC_a meets B_aC_b at NC_aA_b meets C_bA_c at P. Then M,N,P are collinear (Desargues's theorem). M,N,P also lie on this line. 

Now we have 9 points are collinear

Không có nhận xét nào: