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Problema.(Teorema trisectiei) In triunghiul \( ABC \) fie medianele \( BB\prime \) si \( CC\prime \). Printr-un punct \( T \in (BC) \) se duc paralelele \( TD \) si \( TE \) la medianele \( BB\prime \) respectiv \( CC\prime, (D\in(AB), E\in(AC)) \). Atunci \( BB\prime \) si \( CC\prime \) impart \( DE \) in trei segmente congruente.

ATENTIE, am schimbat notatiile din enuntul initial !

Fie \( \triangle\ ABC \) si mijloacele \( N \) , \( P \) ale lui \( [AC] \) , \( [AB] \) respectiv. Pentru \( M\in (BC) \) definim

\( \left|\ \begin{array}{ccc}
E\in AC & , & ME\ \parallel\ BN\\\\
F\in AB & , & MF\ \parallel\ CP\end{array}\ \ \wedge\ \ \begin{array}{c}
X\in BN\ \cap\ EF\\\\
Y\in EF\ \cap\ CP\end{array}\ \right|\ . \) Sa se arate ca \( FX\ =\ XY\ =\ YE \) .

A cam trecut "neobservata" aceasta frumoasa problema. Chiar merita sa fie numita "teorema trisectiei".

Demonstratie. Aplicam teorema Menelaus transversalelor/triunghiurilor mentionate corespunzator :

\( \left|\ \begin{array}{cccc}
\overline {BNX}/AEF\ : & \frac {BF}{BA}\cdot\frac {NA}{NE}\cdot\frac {XE}{XF}=1 & \Longrightarrow & \frac {XE}{XF}=\frac {BA}{BF}\cdot\frac {NE}{NA}\\\\\\\\\\

\overline {CPY}/AEF\ : & \frac {CE}{CA}\cdot\frac {PA}{PF}\cdot\frac {YF}{YE}=1 & \Longrightarrow & \frac {YE}{YF}=\frac {CE}{CA}\cdot\frac {PA}{PF}\end{array}\ \right|\ . \) Se observa usor ca :

\( \begin{array}{c}

\left|\ \begin{array}{c}
\frac {BA}{BF}=2\cdot\frac {BP}{BF}=2\cdot \frac {BC}{BM}\\\\\\\\
\frac {NE}{NA}=\frac {NE}{NC}=\frac {BM}{BC}\end{array}\ \right|\ \Longrightarrow\ \frac {BA}{BF}\cdot \frac {NE}{NA}=2\ \Longrightarrow\ \frac {XE}{XF}=2\\\\\\\\\\\\\\\\\\\\
\left|\ \begin{array}{c}
\frac {CE}{CA}=\frac 12\cdot\frac {CE}{CN}=\frac 12\cdot \frac {CM}{CB}\\\\\\\\
\frac {PA}{PF}=\frac {PB}{PF}=\frac {CB}{CM}\end{array}\ \right|\ \Longrightarrow\ \frac {CE}{CA}\cdot \frac {PA}{PF}=\frac 12\ \Longrightarrow\ \frac {YE}{YF}=\frac 12\end{array}\ \Longrightarrow\ FX\ =\ XY\ =\ XE\ . \)

Generalizare.

Virgil Nicula wrote:Fie triunghiul \( ABC \) si punctele \( \left|\ \begin{array}{ccc}
N\in (AC) & , & \frac {NA}{NC}=n\\\\\\\
P\in (AB) & , & \frac {PA}{PB}=p\end{array}\ \right|\ . \) Pentru \( M\in (BC) \) definim punctele

\( \left|\ \begin{array}{ccc}
E\in AC & , & ME\ \parallel\ BN\\\\
F\in AB & , & MF\ \parallel\ CP\end{array}\ \ \wedge\ \ \begin{array}{c}
X\in BN\ \cap\ EF\\\\
Y\in EF\ \cap\ CP\end{array}\ \right|\ \Longrightarrow\ XY\ =\ \frac {FX}{n}\ =\ \frac {YE}{p}\ =\ \frac {EF}{n+p+1} \) .

Alta solutie:
\( \mbox{Folosind notatiile initiale si punand: } \{M\}=BB^{\prime}\cap DE; \{N\}=CC^{\prime}\cap DE\mbox{, avem de aratat ca: } |DM|=|MN|=|NE|.\\
1).\mbox{Daca: } \{G\}=BB^{\prime}\cap CC^{\prime} \mbox{, atunci }\\
\frac{|GB^{\prime}|}{|GB|}=\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}. \ (1)\\
2).\mbox{Notand cu } \{P\}=DT\cap BB^{\prime}\mbox{, avem:}\\
\left \begin{\array} {\frac{|GC^{\prime}|}{|GC|}=\frac{1}{2}}\ {(1)} \\
{DT||CC^{\prime}} \right\}\Rightarrow \frac{|DP|}{|PT|}=\frac{1}{2}.\ (2) \\
3). \\

\left \begin{\array} {\frac{|DP|}{|PT|}=\frac{1}{2}}\ {(2)} \\
{TE||PM} \right\}\Rightarrow \frac{|DM|}{|ME|}=\frac{1}{2}\Rightarrow \frac{|DM|}{|DE|}=\frac{1}{3}.\ (3) \\
\mbox{In mod analog aratam ca: } \frac{|NE|}{|DE|}=\frac{1}{3}. \ (4)\\
\mbox{In fine, din (3) si (4) }\Rightarrow |DM|=|MN|=|NE|. \)