Đáp án:

$B.\ 45^\circ$

Giải thích các bước giải:

Xét $\triangle ABM$ và $\triangle BCN$ có:

$\begin{cases}AB = BC\\BM = CN\\\widehat{B}=\widehat{C}=90^\circ\end{cases}$

Do đó $\triangle ABM=\triangle BCN\ (c.g.c)$

$\Rightarrow \widehat{BAM}=\widehat{CBN}$ (hai góc tương ứng)

mà $\widehat{CBN}+\widehat{ABN}= \widehat{A}= 90^\circ$

nên $\widehat{BAM} +\widehat{ABN}= 90^\circ$

$\Rightarrow AM\perp BN$

$\Rightarrow \dfrac12\left(sđ\mathop{AB}\limits^{\displaystyle\frown} + sđ\mathop{EF}\limits^{\displaystyle\frown}\right)= 90^\circ$

$\Rightarrow sđ\mathop{AB}\limits^{\displaystyle\frown} + sđ\mathop{EF}\limits^{\displaystyle\frown} = 180^\circ$

$\Rightarrow \widehat{AOB}+ 2\widehat{EDF}= 180^\circ$

$\Rightarrow \widehat{EDF}=\dfrac{180^\circ - \widehat{AOB}}{2}$

$\Rightarrow \widehat{EDF}=\dfrac{180^\circ - 90^\circ}{2}$

$\Rightarrow \widehat{EDF}= 45^\circ$