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O.VI.31

 
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Andi Brojbeanu
Newton


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

PostPosted: Sun Apr 19, 2009 6:28 pm    Post subject: O.VI.31 Reply with quote

Se considera triunghiul isoscel ABC(AB=AC). Punctul E este situat pe perpendiculara in B pe AB, iar punctul F este situat pe perpendiculara in C pe AC, astfel incat E si F sunt in exteriorul unghiului BAC, iar BE=CF. Fie \{O\}=BF\cap CE si \{D\}=BE\cap CF.
Sa se arate ca BF=CE; OB=OC; DB=DC.
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salazar
Thales


Joined: 06 Apr 2009
Posts: 100
Location: Alba Iulia

PostPosted: Sun Apr 26, 2009 7:16 am    Post subject: Reply with quote

\triangle ABC isoscel \Longrightarrow m(\angle ABC)=m(\angle ACB)(1)
m(\angle EBC)=90+m(\angle ABC)(2)
m(\angle FCB)=90+m(\angle ACB)(3)
-din (1),(2),(3)\Longrightarrow \angle EBC\equiv \angle FCB
\triangle EBC\equiv\triangle FCB\Longrightarrow BF=CE, \angle ECB\equiv\angle FBC
\angle OCB\equiv \angle OBC(E-O-C,F-O-B)\Longrightarrow \triangle OBC isoscel, OB=OC.
\angle DBC=180-\angle EBC(4)
\angle DCB=180-\angle FCB(5)
-din (3),(4),(5)\Longrightarrow \angle DBC\equiv \angle DCB\Longrightarrow \triangle DBC isoscel, DB=DC.
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