\( \left|\begin{array}{c}
\triangle ABC,\ M\in\left(AB\right),\ N\in\left(AC\right)\\
BM=CN\\
Q\in\left(MN\right),\ R,\ X\in BC\\
MQ=QN,\ \angle BAX=\angle CAX,\ BR=RC\end{array}\right|\Longrightarrow QR\parallel AX \)
Dreapta paralela cu bisectoarea
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Claudiu Mindrila
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Dreapta paralela cu bisectoarea
Last edited by Claudiu Mindrila on Sat Jun 27, 2009 12:50 pm, edited 1 time in total.
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Mainea
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Folosim proprietatea \( \overline{QR}=\frac{1}{2}\(\overline{MB}+\overline{NC}\) \).
De aici rezulta totul, deoarece \( \overline{MB} \) si \( \overline{NC} \) avand acelasi modul rezultanta lor are directia bisectoarei aceestora.
De aici rezulta totul, deoarece \( \overline{MB} \) si \( \overline{NC} \) avand acelasi modul rezultanta lor are directia bisectoarei aceestora.
Last edited by Marius Mainea on Sat Jun 27, 2009 12:51 pm, edited 1 time in total.
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Mateescu Constantin
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Solutie sintetica:
Fie \( MM^{\prime}\ \parallel\ AX\ \parallel\ NN^{\prime},\ M^{\prime},\ N^{\prime}\in[BC] \) .
\( \Longrightarrow\ \left\|\begin{array}{cc} \frac{BM^{\prime}}{BX} & = & \frac{BM}{BA} & (\sim) & \Longleftrightarrow & \frac{AB}{BX} & = & \frac{BM}{BM^{\prime}} \\\\\\\\
\frac{CN^{\prime}}{CX} & = & \frac{CN}{CA} & (\sim) & \Longleftrightarrow & \frac{AC}{CX} & = & \frac{CN}{CN^{\prime}}
\end{array}\ \right\|\ \ \wedge\ \frac{AB}{BX} = \frac{AC}{CX}\ (\mbox{T.Bis.})\ \Longrightarrow^{\small BM=CN}\ BM^{\prime}=CN^{\prime} \)
Deci \( RM^{\prime}=RN^{\prime} \) si cum \( MQ=QN \) rezulta ca \( [RQ] \) este linie mijlocie in trapezul \( MM^{\prime}N^{\prime}N \).
\( \Longrightarrow\ RQ\ \parallel\ MM^{\prime}\ \parallel\ AX \).
Fie \( MM^{\prime}\ \parallel\ AX\ \parallel\ NN^{\prime},\ M^{\prime},\ N^{\prime}\in[BC] \) .
\( \Longrightarrow\ \left\|\begin{array}{cc} \frac{BM^{\prime}}{BX} & = & \frac{BM}{BA} & (\sim) & \Longleftrightarrow & \frac{AB}{BX} & = & \frac{BM}{BM^{\prime}} \\\\\\\\
\frac{CN^{\prime}}{CX} & = & \frac{CN}{CA} & (\sim) & \Longleftrightarrow & \frac{AC}{CX} & = & \frac{CN}{CN^{\prime}}
\end{array}\ \right\|\ \ \wedge\ \frac{AB}{BX} = \frac{AC}{CX}\ (\mbox{T.Bis.})\ \Longrightarrow^{\small BM=CN}\ BM^{\prime}=CN^{\prime} \)
Deci \( RM^{\prime}=RN^{\prime} \) si cum \( MQ=QN \) rezulta ca \( [RQ] \) este linie mijlocie in trapezul \( MM^{\prime}N^{\prime}N \).
\( \Longrightarrow\ RQ\ \parallel\ MM^{\prime}\ \parallel\ AX \).
Last edited by Mateescu Constantin on Tue Aug 18, 2009 3:10 pm, edited 4 times in total.
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maxim bogdan
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