Respostas

2013-10-19T20:21:22-03:00
BAC\sim{GEF}
\dfrac{{\frac{x}{2}}}{GF}=\dfrac{x}{EF}

EF=2GF
GF=\dfrac{EF}{2}


Pitágoras
GEF 
(GE)^2=(EF)^2+\left(\dfrac{EF}{2}^2\right)

GE=\dfrac{EF\sqrt{5}}{2}

Triângulo ABG Pitágoras
(BG)^2=x^2+\left(\dfrac{x}{2}\right)^2

BG=\dfrac{x\times\sqrt{5}}{2}


Relação entre cordas
BG\times{GE}=\dfrac{x}{2}\times\dfrac{x}{2}

\dfrac{x\times\sqrt{5}}{2}\times\dfrac{{EF}\times\sqrt{5}}{2}=\dfrac{x^2}{4}
x=5EF
\dfrac{x}{2}=AG=\dfrac{5EF}{2}

Mas,  
AF=AG+GE
AF=\dfrac{5EF}{2}+\dfrac{EF}{2}

AF=\dfrac{6EF}{2}=3EF