Giải thích các bước giải:
a.Vì $\Delta ABC,\hat{A}=90^o,\hat{C}=45^o\rightarrow\Delta ABC$ vuông cân tại A
$\rightarrow\widehat{BAD}=\dfrac{1}{2}\widehat{BAC}=45^o\rightarrow\widehat{EAB}=180^o-\widehat{BAD}=135^o$
Lại có : $\widehat{BCF}=180^o-\widehat{ACB}=135^o$
$\rightarrow\widehat{BAD}=\widehat{ACB}$
Mà $AE=BC, AB=AC=CF\rightarrow \Delta AEB=\Delta CBF(c.g.c)$
b.Từ câu a $\rightarrow\widehat{EBA}=\widehat{CFB}$
$\rightarrow\widehat{EBF}=\widehat{EBA}+\widehat{ABC}+\widehat{CBF}=\widehat{CFB}+\widehat{ABC}+\widehat{CBF}=\widehat{CFB}+\widehat{CBF}+\widehat{ABC}=\widehat{ACB}+\widehat{ABC}=90^o$
$\rightarrow BE\perp BF$