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Un trapez si doua cercuri.
Posted: Mon May 26, 2008 8:31 pm
by Virgil Nicula
Fie trapezul \( ABCD \) , unde \( AB\parallel CD \) . Notam \( O\in AC\cap BD \) . Cercurile circumscrise triunghiurilor \( AOD \) , \( BOC \) se taie
din nou in \( K \) . Notam \( S\in KO\in AB \) . Sa se arate ca \( \frac {SA}{SB}=\left(\frac {OA}{OB}\right)^2 \) (dreapta \( \overline {OKS} \) este simediana in triunghiul \( AOB \) ).
Posted: Wed May 28, 2008 8:41 pm
by Marius Mainea
\( \triangle \)\( BOS \)\( \sim \)\( \triangle \)\( DCB \) \( \Rightarrow \) \( \frac{SB}{BD} \)=\( \frac{OB}{DC} \) (1).
\( \triangle \)\( AOS \)\( \sim \)\( \triangle \)\( CDA \)\( \Rightarrow \) \( \frac{SA}{AC} \)=\( \frac{OA}{DC} \) (2).
\( \frac{OA}{AC} \)=\( \frac{OB}{BD} \) (3).
Din (1) ,(2) si (3) rezulta concluzia.
Posted: Thu May 29, 2008 8:42 am
by Virgil Nicula
Asemanarile tale sunt adevarate doar daca trapezul este paralelogram ...
Posted: Thu May 29, 2008 9:11 am
by Marius Mainea
Aveti dreptate.
Posted: Thu May 29, 2008 9:50 am
by Marius Mainea
Fie E si F intersectiile cercurulor AOD respectiv BOC co dreapta AB.
Folosind puterea punctului fata de cerc avem:
AO\( \cdot \)AC=AF\( \cdot \)AB (1)
BO\( \cdot \)BD=BE\( \cdot \)AB (2)
Deasemenea\( \frac{OA}{AC} \)=\( \frac{OB}{BD} (3) \)
Din1 ,2,3 rezult: \( (\frac{OA}{OB})^2 \)=\( \frac{AF}{BE} \)
Dar S are aceeasi putere fata cele doua cercuri\( \Rightarrow \)SE\( \cdot \)SA=SF\( \cdot \)SB\( \Rightarrow \)\( \frac{AF}{BE}=\frac{SA}{SB} \)
Rezulta concluzia problemei.