IME / ITA ⇒ (CN) Semelhança Tópico resolvido

[castelohsi]

Um triângulo ABC está inscrito num círculo e os lados AB e BC medem, respectivamente, 15cm e 25cm. A reta traçada pelo vértice A, paralela à tangente ao círculo no ponto B, intercepta o lado BC no ponto D. Calcule a medida de DC.

Resposta

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16cm

[petras]

castelohsi,

Distribuindo os ângulos você encontrará a semelhança dos triângulos

[tex3]\mathtt{Seja~\angle FBA(a.segmento)=\frac{\theta}{2}\implies \angle BOA = \theta~e~\angle BCA = \frac{\theta}{2}\\
\angle ABO=\angle BAO = 90-\frac{\theta}{2}\\
\angle ABC = \frac{180 -\frac{\theta}{2}}{2}=90-\frac{\theta}{4} \\
\angle ADC = \angle FBD = \frac{\theta}{2}+90-\frac{\theta}{4}=90+\frac{\theta}{4}\implies \angle CAD = 90-\frac{3\theta}{4}\\
\angle DAB = 90-\frac{\theta}{4}-(90-\frac{3\theta}{4} = \frac{\theta}{2}\\
\therefore \triangle ABC \sim \triangle CAD(A.A.)\implies \frac{BD}{AB} = \frac{AB}{BC}\implies BD =\frac{225}{25}=9\\
\therefore CD = 25-9 = \boxed{16\color{green}}\checkmark

}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathtt{Seja~\angle FBA(a.segmento)=\frac{\theta}{2}\implies \angle BOA = \theta~e~\angle BCA = \frac{\theta}{2}\\
\angle ABO=\angle BAO = 90-\frac{\theta}{2}\\
\angle ABC = \frac{180 -\frac{\theta}{2}}{2}=90-\frac{\theta}{4} \\
\angle ADC = \angle FBD = \frac{\theta}{2}+90-\frac{\theta}{4}=90+\frac{\theta}{4}\implies \angle CAD = 90-\frac{3\theta}{4}\\
\angle DAB = 90-\frac{\theta}{4}-(90-\frac{3\theta}{4} = \frac{\theta}{2}\\
\therefore \triangle ABC \sim \triangle CAD(A.A.)\implies \frac{BD}{AB} = \frac{AB}{BC}\implies BD =\frac{225}{25}=9\\
\therefore CD = 25-9 = \boxed{16\color{green}}\checkmark

}[/tex3]