Estude! Com Prof. Caju

Na figura abaixo, AB//EF//CD, AB = 10 e CD = 6. Sabendo disso, calcule o valor numérico de EF.

FISMAQUIM,

Seja BF = x, CF = y e EF = m

[tex3]\triangle CEF \sim \triangle CAB: \frac{m}{y} = \frac{10}{x+y}\implies m = \frac{10y}{x+y}(I)\\
\triangle BEF \sim \triangle BDC: \frac{m}{x} = \frac{6}{x+y}\implies m = \frac{6x}{x+y}(II)\\
Igualando:10y = 6x \implies x =\frac{5y}{3} \therefore BC = \frac{5y}{3}+y = \frac{8y}{3}=x+y\\
De(I): m =\frac{10y}{\frac{8y}{3}}=\frac{30}{8} = \boxed{\frac{15}{4}}
[/tex3]

petras, muito obrigado