Capítulo 5 ⇒ Solucionário:Racso - Cap V - Problemas de Geometria y Como Resolverlos - I Edição - Ex:08 Tópico resolvido

[petras]

Problema Proposto
8 - Na figura se sabe que:
AP = PB, AN = NC e BC= 6[tex3]\sqrt{2}[/tex3].
Calcular MN

Resposta

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E) 3[tex3]\sqrt{2}[/tex3]

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[tex3]\mathsf{
Traçar ~PN \perp AC \implies \triangle APC ~é~isósceles\\
\angle APN = \angle APC = \frac{180-\theta}{2}=90-\frac{\theta}{2}\\
\therefore \angle NPC = \angle NPA = 180-90-(90-\frac{\theta}{2})=\frac{\theta}{2}\\
\frac{\theta}{2}+\frac{\theta}{2}+\theta = 180^o\\
\therefore \theta = 60^o \implies \triangle ABC~é~equilátero \\
MN = \frac{BC}{2} = \boxed{\color{red}3\sqrt2}
}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
Traçar ~PN \perp AC \implies \triangle APC ~é~isósceles\\
\angle APN = \angle APC = \frac{180-\theta}{2}=90-\frac{\theta}{2}\\
\therefore \angle NPC = \angle NPA = 180-90-(90-\frac{\theta}{2})=\frac{\theta}{2}\\
\frac{\theta}{2}+\frac{\theta}{2}+\theta = 180^o\\
\therefore \theta = 60^o \implies \triangle ABC~é~equilátero \\
MN = \frac{BC}{2} = \boxed{\color{red}3\sqrt2}
}[/tex3]