Triângulos - 2008 - Vol. 2 ⇒ 052 - Triângulos - 2008 Tópico resolvido

[petrasMOD]

Na figura, calcular "x".

Resposta

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Gabarito: 45^o

[rcompany]

[tex3]\triangle ABC\text{ o triângulo}\\
\widehat{A}=x,\,\widehat B=\gamma+2\varphi,\,\widehat C=\varphi+2\gamma\\
BM, CN \text{ as cevianas conforme a figura}\\
O\text{ a intersecção dessas duas cevianas}\\
\angle BOC=\angle MON\implies \pi-2\gamma -2\varphi=\pi-\angle OMN-\angle ONM=2\theta+2\beta-\pi\implies \beta+\gamma+\theta+\varphi=\pi\\
\angle COM=\angle NOB\implies \pi-\theta-\varphi=\pi-\beta-\gamma\implies \beta+\gamma=\theta+\varphi\\
\therefore \beta+\gamma=\theta+\varphi=\frac{\pi}{2}\\
P\text{ o ponto em $NB$ tal que $CP$ seja bissetriz de $\angle NCB$}\\
\beta+\gamma=\frac{\pi}{2}\implies \angle CPN=\frac{\pi}{2}\\
CP\text{ bissetriz e altura}\implies \triangle NCB\text{ isósceles}\implies \angle PNC=\angle CBP\implies \beta=\gamma+2\varphi\\
\text{(Da mesma forma temos } \theta=\varphi+2\gamma)\\
\beta=\gamma+2\varphi\text{ e }\beta+\gamma=\frac{\pi}{2}\implies \phi+\gamma=\frac{\pi}{4}\\
x=\angle PAC=\pi-\angle CPA-\angle ACP=\pi-\frac{\pi}{2}-(\varphi+\gamma)=\pi-\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}=45°

[/tex3]

[petrasMOD]

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