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Laturile opuse ale hexagonului convex ABCDEF sunt paralele doua cate doua. Sa se arate ca ariile triunghiurilor ACE si BDF sunt egale, valoarea lor comuna fiind cel putin jumate din aria hexagonului.

\( \mbox{Prin varfurile A, B, C, D, E si F ducem dreptele a, b, c, d, e si respectiv f, astfel incat sa avem: } \)
\( a\parallel d\parallel BC\parallel EF; \ \ b\parallel e\parallel AF\parallel f CD; \ \ c\parallel f\parallel AB\parallel DE \ \ \mbox{ si notam apoi cu: } \)
\( \{A_1}=c\cap e; \{C_1\}=a\cap e; \{E_1\}=a\cap c; \{B_1\}=d\cap f; \{D_1\}=b\cap f; \{F_1\}=b\cap d. \)
\( \mbox{Sa observam acum ca patrulaterele: }ABCE_1, CDEA_1, EFAC_1 \mbox{-fiind paralelograme}\Rightarrow \)
\( \Rightarrow S_{ACE}=S_{ACE_1}+S_{CEA_1}+S_{AEC_1}+S_{A_1C_1E_1}=\frac{1}{2}.S_{ABCDEF}+ S_{A_1C_1E_1}\ge \frac{1}{2}.S_{ABCDEF}. \) (1)
\( \mbox{In mod analog se arata ca: } S_{BDE}=\frac{1}{2}.S_{ABCDEF}+ S_{B_1D_1F_1}\ge \frac{1}{2}.S_{ABCDEF}. \) (2)
\( \mbox{Pe baza relatiilor (1) si (2), avem: } S_{ACE}=S_{BDF}\Leftrightarrow S_{A_1C_1E_1}=S_{B_1D_1F_1}. \) (3)
\( \mbox{Avem insa: }\\

A_1C_1=|AF-CD|=D_1F_1; C_1E_1=|BC-EF|=B_1F_1; A_1E_1=|AB-DE|=B_1D_1\Rightarrow S_{A_1C_1E_1}=S_{B_1D_1F_1}.
\) (3)

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