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Doua probleme interesante de geometrie

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

PostPosted: Sat Jun 28, 2008 2:21 pm    Post subject: Doua probleme interesante de geometrie Reply with quote

1. Pe segmentul (AB) se considera un punct P. De aceeasi parte a dreptei AB se construiesc triunghiurile \Delta APC si \Delta APD astfel incat [PA] \equiv [PC], [PB] \equiv [PD] si m(\angle APC)=m(\angle BPD)=a. Dreptele AD si  BC se intersecteaza in Q. Sa se determine m(\angle AQC) in functie de a.

Maria Mihet, R.M.T. 2/2008

2. Se considera triunghiul \Delta ABC, dreptunghic in A. Sa se arate ca unul din unghiurile B si C este egal cu 30^ \circ daca si numai daca AI=MI unde I este centrul cercului inscris triunghiului, iar M mijlocul ipotenuzei.

Mihail Mogosanu, R.M.T. 2/2008
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Andi Brojbeanu
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Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Sat Apr 18, 2009 3:37 pm    Post subject: Reply with quote

Problema 1. Sunt 3 posibilitati:
1) a<90\textdegree. m(\widehat{APD})=m(\widehat{BPC})(=a+m(\widehat{CPD})), deci  \bigtriangleup {APD}\equiv\bigtriangleup{CPB}(L.U.L). Obtinem ca m(\widehat{BCP})=m(\widehat{DAP}). Deci, m(\widehat{ACQ})=m(\widehat{QAB})+m(\widehat{QBA)}=m(\widehat{BCP})+m(\widehat{CBP})=m(\widehat{CPA})=a.
2) a=90\textdegree. Deoarece Q este punctul unde dreapta BC taie dreapta AD, inseamna ca m(\widehat{BCQ})=180\textdegree. Atunci m(\widehat{ACQ})=m(\widehat{BCQ})-m(\widehat{ACP})-m(\widehat{BCP})=180\textdegree-45\textdegree-45\textdegree=90\textdegree.
3) a>90\textdegree. Insa m(\widehat{APD})=m(\widehat{CPB})(=a-m(\widehat{CPD})), rezulta \bigtriangleup{APD}\equiv\bigtriangleup{CPB}(L.U.L). Analog cazului anterior, m(\widehat{BCP})=m(\widehat{DAP}).
m(\widehat{ACQ})=180\textdegree-m(\widehat{QAB})-m(\widehat{QBA})=180\textdegree=m(\widehat{BCP})-m(\widehat{CPB})=m(\widehat{CPB})=180\textdegree-a.

Problema 2. "\Rightarrow" Daca m(\widehat{B})=30\textdegree, rezulta AC=\frac{BC}{2}=CM. Deci, \bigtriangleup{AIC}\equiv\bigtriangleup{MIC}(L.U.L) rezulta AI=MI.
"\Leftarrow" Luam D si E proiectiile luiI pe [AC], respectiv BC. Evident, ID=IE, deci \bigtriangleup{AID}\equiv \bigtriangleup{MIE}(L.U.L).
Daca AB\leq AC, avem \widehat{IMP}\equiv\widehat{IMB}\equiv\widehat{IAN}\equiv\widehat{CAI}\equiv\widehat{BAI}. Rezulta \bigtriangleup{AIB}\equiv\bigtriangleup{MIB}(U.L.U), adica AB=BM=\frac{BC}{2} , de unde m(\widehat{C})=30\textdegree.
Analog, pentru AB>AC, AC=MC=\frac{BC}{2}, rezulta m(\widehat{B})=30\textdegree.
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