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OLM Timis 2002

 
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alex2008
Leibniz


Joined: 19 Oct 2008
Posts: 489
Location: Tulcea

PostPosted: Wed Nov 12, 2008 7:26 am    Post subject: OLM Timis 2002 Reply with quote

Fie unghiul propriu AOD si semidreptele [OB si [OC in interiorul sau astfel incat [OB sa fie interiorul unghiului AOC si unghiul AOB congruent cu unghiul COD . Stiind ca [AO] congruent cu [OD], [BO] congruent [OC] , iar M si N sunt mijloacele segmentelor AB, respectiv CD, aratati ca [DM] este congruent cu [AN].



Problema foarte simpla !
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Andi Brojbeanu
Newton


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

PostPosted: Fri Apr 17, 2009 10:22 pm    Post subject: Reply with quote

Stiind din ipoteza ca AO=OD, BO=CO si \widehat{AOB}\equiv\widehat{COD}, deducem ca triunghiul AOB congruent cu triunghiul COD (L.U.L)\RightarrowAB=CD si CN=BM.
Atunci CD+BC+BM=AB+BC+CN, deci DM=AN.
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