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.