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O problema draguta de geometrie

 
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Claudiu Mindrila
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Joined: 01 Oct 2007
Posts: 557
Location: Targoviste

PostPosted: Tue Apr 27, 2010 3:57 pm    Post subject: O problema draguta de geometrie Reply with quote

Fie triunghiul ABC in care \angle C=60^{\circ}. Pe prelungirea laturii AC dincolo de C se ia punctul D, iar pe prelungirea laturii BC dincolo de C se ia punctul E astfel incat BD=DE. Daca AD=CE demonstrati ca triunghiul ABC este echilateral.

Calin Burdusel, Olimpiada Judeteana de Matematica, Dambovita, 2010
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Andi Brojbeanu
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Joined: 22 Mar 2009
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PostPosted: Wed May 12, 2010 5:05 pm    Post subject: Reply with quote

Deoarece BD=DE, triunghiul BDE este isoscel.
Fie F mijlocul segmentului [BE]. Atunci, DF mediana, deci si inaltime.
In triunghiul dreptunghic CDF, masura unghiului \angle{DCF}(\equiv \angle {ACB}) este egala cu 60\textdegree, deci masura unghiului \angle{CDF} este 30\textdegree. Conform teoremei unghiului de 30\textdegree, CD=2CF.
Din AD=CE\Rightarrow AC+CD=CE\Rightarrow AC+2CF=CF+FE\Rightarrow AC=FE+CF-2CF=FE-CF=BF-CF=BC\Rightarrow \bigtriangleup{ABC} este isoscel.
Deoarece m(ACB)=60\textdegree \Rightarrow \bigtriangleup{ABC} este echilateral.
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Claudiu Mindrila
Fermat


Joined: 01 Oct 2007
Posts: 557
Location: Targoviste

PostPosted: Wed May 12, 2010 6:15 pm    Post subject: Reply with quote

Uite cum m-am gandit eu:

Fie M al patrulea varf al paralelogramului DECM. Cum DM=EC=DA si \widehat{ADM}=\widehat{C}=60^{\circ} rezulta ca \triangle{ADM} este echilateral.
Deoarece \triangle BDM\equiv\triangle DEC\ (LUL) rezulta ca \widehat{DMB}=\widehat{ECD}=60^{\circ}.
Dar \widehat{DMB}=\widehat{DMA}=60^{\circ}\Longrightarrow M\in AB\Longrightarrow\widehat{A}=60^{\circ}, c.c.t.d.
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