Circunferência - 2003 - Vol. 5 ⇒ Problema 30 - Circunferência -Vol. 5 Tópico resolvido

[petrasMOD]

Se AB = 9 e FC = 3 calcule BC.

Resposta

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Gabarito: B) 6

[geobson]

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[rcompany]

@geobson
"Introdução ao Latex: um Minicurso para Iniciantes"
https://revista.gru.ifsp.edu.br/semat/article/view/207

[petrasMOD]

[tex3] \overset{\LARGE{\frown}}{CM}= \overset{\LARGE{\frown}}{MD} \implies \\
\angle MAC \cong \angle DAM = \theta\\
\therefore ACMD: \angle DCM \cong \angle DAM = \theta\\
\angle MAC \cong \angle MDC = \theta\\
\angle CDA = \alpha \implies \overset{\LARGE{\frown}}{AC} = 2\alpha\\
\angle BAC_{(semiinscrito)} = \frac{ \overset{\LARGE{\frown}}{AC}}{2} = \alpha\\
\therefore \angle FAB = \theta +\alpha(i)\\
\angle CFA = \alpha + \theta(ii)\\
De(i)e(ii): \triangle ABF_{(isosc)} \implies AB = BF = 9\\
\therefore BC = BF - FC = 9-3 = \boxed{6}



[/tex3]