Se considera patrulaterul convex \( ABCD \) in care \( \angle A=60^{\circ},\ \angle C=30^{\circ} \) si \( AB=AD \).
Aratati ca \( AC^2=BC^2+CD^2 \) .
Relatie intr-un patrulater convex
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Mateescu Constantin
- Newton
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Marius Mainea
- Gauss
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Din relatia lui Euler \( AC^2+BD^2=AB^2+BC^2+CD^2+DA^2-4EF^2 \) obtinem ce relatia din concluzie este echivalenta cu
\( EF=\frac{AB}{2} \)
\( EF=\frac{AB}{2} \)
Last edited by Marius Mainea on Fri Aug 14, 2009 10:12 pm, edited 2 times in total.
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Mateescu Constantin
- Newton
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Demonstratie:
Aplicam teorema lui Ptolemeu generalizata:
\( AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+AD^{2}\cdot BC^{2}-2AB\cdot BC\cdot CD\cdot DA\cdot \cos(A+C) \)
\( \stackrel{A+C=90^{\circ}}{\Longleftrightarrow }AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+AD^{2}\cdot BC^{2}\ \Longleftrightarrow\ AC^{2}=CD^{2}+BC^{2} \) , deoarece triunghiul \( ABD \) este echilateral.
Aplicam teorema lui Ptolemeu generalizata:
\( AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+AD^{2}\cdot BC^{2}-2AB\cdot BC\cdot CD\cdot DA\cdot \cos(A+C) \)
\( \stackrel{A+C=90^{\circ}}{\Longleftrightarrow }AC^{2}\cdot BD^{2}=AB^{2}\cdot CD^{2}+AD^{2}\cdot BC^{2}\ \Longleftrightarrow\ AC^{2}=CD^{2}+BC^{2} \) , deoarece triunghiul \( ABD \) este echilateral.