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Problema Proposto
43 - No hexágono regular ABCDEF se traça em sua região interior o quadrado AFGH
Sobre EF se marca o ponto M; se CM intercepta em N a GF, calcular CN,
se MN = 1 e m [tex3]\angle NGM = 30^o[/tex3]

[tex3]Traçar ~ JM \perp FG~and ~FD \parallel AH \\
\angle JMC = \angle MCD(ângulos ~alternos )\\
\angle MGN = \angle IFM = 30^o \therefore \triangle FMG (isosceles)\implies MF = MG\\
a =lado~AFGH\\\triangle FMI:
MI = \frac{a\sqrt3}{6}\\
\triangle IMN \sim \triangle DCN \implies\frac{IM}{CD} = \frac{1}{CN} = \frac{DN}{IN}\\
\frac{CN}{1} = \frac{a}{\frac{a\sqrt3}{6}}=\frac{6}{\sqrt3} \\
\therefore \boxed{\color{red} CN = 2\sqrt3}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]Traçar ~ JM \perp FG~and ~FD \parallel AH \\
\angle JMC = \angle MCD(ângulos ~alternos )\\
\angle MGN = \angle IFM = 30^o \therefore \triangle FMG (isosceles)\implies MF = MG\\
a =lado~AFGH\\\triangle FMI:
MI = \frac{a\sqrt3}{6}\\
\triangle IMN \sim \triangle DCN \implies\frac{IM}{CD} = \frac{1}{CN} = \frac{DN}{IN}\\
\frac{CN}{1} = \frac{a}{\frac{a\sqrt3}{6}}=\frac{6}{\sqrt3} \\
\therefore \boxed{\color{red} CN = 2\sqrt3}
[/tex3]