## Thứ Bảy, 13 tháng 9, 2014

### 57-A generalization Newton's line theorem

Let the line (d) meets three sides of the triangle at $A_1,B_1,C_1$. Denote $A_2,B_2,C_2$ are midpoints of $AA_1,BB_1,CC_1$ then $A_2,B_2,C_2$ are collinear. The line through $A_2,B_2,C_2$ call Newton's line

Theorem(A generalization Newton line theorem): Let $P$ be a point on the plain,$EFG$ be cevian triangle of the point P. The line AA_1,BB_1,CC_1 meets three sidelines of $EFG$ at $A'_2, B'_2,C'_2$ then $A'_2, B'_2,C'_2$ are collinear.

-  When $P$ is the centroid of $ABC$, this theorem is Newton line theorem above
Solution by Telv Cohl (Telv Cohl: National Chiayi Senior High School, Chiayi, Taiwan. E-mail address: telvcohltinasprout@gmail.com)
Proof 1 :
$X$ = the intersection of $AE$ and $GF$
$Y$ = the intersection of $BF$ and $GE$
$Z$ = the intersection of $CG$ and $EF$

By Ceva theorem we get $\frac{AG}{GB}.\frac{BE}{EC}.\frac{CF}{FA}=1$                                                      (1)

By Menelaus theorem we get $\frac{CA_1}{A_1B}.\frac{BC_1}{C_1A}.\frac{AB_1}{B_1C}=1$                        (2)

By Ceva theorem we get $\frac{GX}{XF}.\frac{FZ}{ZE}.\frac{EY}{YG}=1$                                                      (3)

Since $(B,C;E,A_1)=(G,F;X,A_2)$ we get $\frac{BE}{EC}.\frac{CA_1}{BA_1}=\frac{GX}{XF}.\frac{FA_2}{A_2G}$         (4)

Since $(C,A;F,B_1)=(E,G;Y.B_2)$ we get $\frac{CF}{FA}.\frac{CB_1}{AB_1}=\frac{EY}{YG}\frac{GB_2}{B_2F}$             (5)

Since $(A,B;G,C_1)=(F,E;Z,C_2)$ we get $\frac{AG}{GB}.\frac{BC_1}{AC_1}=\frac{FZ}{ZE}.\frac{EC_2}{C_2F}$           (6)

From (1) (2) (3) (4) (5) (6) we get $\frac{FA_2}{A_2G}.\frac{GB_2}{B_2E}.\frac{EC_2}{C_2F}=1$

By the converse of Menelaus theorem we get $A_2, B_2, C_2$ are collinear
Q.E.D
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