In triangle \(ABC\), angle A measures \(35\degree\), angle \(B\) measures \(75 \degree\), and angle \(C\) measaures \(70 \degree\). Triangle \(DEF\) is similar to triangle \(ABC\), such that \(\dfrac{DE}{AB}=\dfrac{EF}{BC}=\dfrac{DF}{AC}=2 \). What is the measure, in degrees, of angle \(E\)?

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$$\dfrac{D\colorbox{aqua}{$E$}}{A\colorbox{aqua}{$B$}}=\dfrac{\colorbox{aqua}{$E$}F}{\colorbox{aqua}{$B$}C}=\dfrac{DF}{AC}=2 $$

If we compare corresponding vertices, \(E\) corresponds to \(B\). Because the triangles are similar,

$$ m\angle E = m \angle B = 75 \degree$$ $$ m \angle E = \boxed{75\degree}$$

Despite triangle \(DEF\) being twice the size of \(ABC\), the angles will still be congruent.