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JBTST III 2007, Problema 1
Posted: Sun Apr 20, 2008 10:32 am
by Laurian Filip
Fie ABC un triunghi si punctele M,N,P pe laturile AB,BC,CA respectiv, astfel incat CPMN sa fie paralelogram. Dreptele AN si MP se intersecteaza in punctul R, dreptele BP si MN se intersecteaza in punctul S, iar Q este punctul de intersectie al dreptelor AN si BP. Sa se arate ca S[MRQS]=S[NQP].
Posted: Tue Jun 30, 2009 7:39 pm
by Mateescu Constantin
Aplicand teorema lui Thales obtinem:
\( \left\|\begin{array}{ccc} \frac{AR}{RN}&=&\frac{AM}{MB}\\\\\\\\\\
\frac{MS}{SN}&=&\frac{SP}{SB}&=&\frac{NC}{NB}\\\\\\\\\\
\frac{AM}{MB}&=&\frac{NC}{NB}\end{array}\right| \ \Longrightarrow\ \frac{AR}{RN}\ =\ \frac{MS}{SN}\ \Longrightarrow^{\mbox{R.T.Th}}\ RS\ \parallel\ AB. \)
Atunci in trapezele \( RPNB \) si \( RSBM \) avem relatiile \( S_{PQN}\ =\ S_{RQB} \) si \( S_{RSB}\ =\ S_{RSM} \), ceea ce conduce la concluzia problemei.