Quadriláteros - 2003 - Vol 4 ⇒ Problema 61 - Quadrilateros -Vol. 4 Tópico resolvido

[petrasMOD]

Na figura calcular x^o.

Resposta

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Gabarito: C) 120^o

[rcompany]

[tex3]
G,H,I\text{ os pontos tais que:}\left\{\begin{array}{l}\angle(AI),(AB)=10°\text{ e }\angle(CB),(CI)=10°\\H \text{ intersecção de }(AI)\text{ e }(BC)\\\angle(AC),(CG)=40°\text{ e }G\text{ intersecção de }(CG)\text{ com }(AD)\end{array}\right.\\

\triangle AIC \equiv \triangle ADC \implies (ID)\perp (AC)\\
\angle AID \equiv \angle ADI=70°\implies \angle DIC\equiv \angle IDC = 60°\\
\angle IHC + \angle IDC=180°\implies HIDC\text{ cíclico de centro }O\\
\angle DIC + \angle DGC=180°\implies IDGC\text{ cíclico de centro }O'\\\\
\left.\begin{array}{rl}EI=ED,EH=EG\\OH=OI=OD=OC\\O'I=O'D=O'G=O'C\end{array}\right\}\implies O=O'=E\\\\
\left.\begin{array}{rl}\angle ICD=60°\implies \angle IED=120°\\\angle ICD=80°\implies \angle HCG=160°\\(AC)\text{ bissetriz de }\angle IED\text{ e }\angle HCG\end{array}\right\}\implies \angle CEB=60°\text{ e }\angle HEA=80°\implies \angle BEH=40°\implies\angle AEB=\angle HEA+\angle BEH=80°+40°=120°
[/tex3]