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Teorema lui Vecten

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Thu May 14, 2009 10:34 am    Post subject: Teorema lui Vecten Reply with quote

Se considera un triunghi ABC. Fie A_1 un punct din plan cu proprietatile:
-dreapta BC separa punctele A si A_1;
-A_1 este centrul patratului construit pe latura [BC].
Analog se obtin punctele B_1 si C_1. Atunci dreptele AA_1, BB_1 si CC_1 sunt concurente. (Punctul de concurenta a celor trei drepte se numeste punctul lui Vecten).
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Thu Jun 18, 2009 6:38 pm    Post subject: Reply with quote

Din asemanarea triunghiurilor ACB_1 si ABC_1 rezulta \frac{AB}{AC}=\frac{AC_1}{AB_1} sau AB\cdot AB_1=AC\cdot AC_1^{(1)}.
Deoarece m(\angle{BAB_1})=m(\angle{CAC_1})\Rightarrow sin \angle{BAB_1}=sin \angle{CAC_1}^{(2)}.
(1) si (2)\Rightarrow AB\cdot AB_1\cdot sin \angle{BAB_1}=AC\cdot AC_1\cdot sin \angle{CAC_1}\Rightarrow \sigma[ABB_1]=\sigma[ACC_1]^{(3)}.
In mod asemanator se obtine \sigma[BCC_1]=\sigma [BAA_1] si \sigma[CAA_1]=\sigma[CBB_1]^{(4)}.
Fie punctele \{A^\prim\}=BC\cap AA_1, \{B^\prim\}=CA\cap BB_1, \{C^\prim\}=AB\cap CC_1.
Se observa ca:
\frac{A^\prim B}{A^\prim C}=\frac{\sigma[BAA_1]}{\sigma[CAA_1]}^{(5)}
\frac{B^\prim C}{B^\prim A}=\frac{\sigma[CBB_1]}{\sigma[ABB_1]}^{(6)}
\frac{C^\prim A}{C^\prim B}=\frac{\sigma[ACC_1]}{\sigma[BCC_1]}^{(7)}
(3), (4), (5), (6), (7)\Rightarrow \frac{A^\prim B}{A\prim C}\cdot \frac{B^\prim C}{B^\prim A}\cdot \frac{C^\prim A}{C^\prim B}=1, din care, conform Reciprocei teoremei lui Ceva rezulta ca dreptele AA_1, BB_1,CC_1 sunt concurente (Teorema lui Vecten).
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