}[/tex3] a)2
b)[tex3]\frac{1}{2}[/tex3]
c)1
d)[tex3]\frac{2}{3}[/tex3]
e)[tex3]\frac{4}{3}[/tex3]
Resposta
c
Moderador: [ Moderadores TTB ]
Show de bola , meu amigo.petras escreveu: ↑Sáb 18 Jun, 2022 13:56geobson,
[tex3]\mathsf{
\angle CEF=\theta \therefore \angle AGB=120^o- θ\\
\angle DBF=\angle DEF=60^o-\theta \implies \angle BCA=60^o ~e~ ACDE ( cíclico) \implies \\
\therefore AC ~e~CD ~"enxergam~o~~mesmo~ arco ~60^o \implies AC = CD. \\
\therefore △ACD (equilátero) \implies AD=AC=CD\\
Trace ~\angle EAH=60^o para ~interceptar~CE ~em ~H.\\
AH=EH=AE \therefore ∠AHC=120^o\\
Como\angle ACH=\angle ADE \implies △ACH≅△ADE\\
\therefore CH=DE ~e~CE=CH+EH=AE+DE \implies \frac{CE}{AE+DE}=
\frac{AE+DE}{AE+DE} = \boxed{1}\color{green}\checkmark
}[/tex3]
(Solução:MathLover)