`a)`Ta có:

$\widehat{ACB}$`+`$\widehat{FCB}$`=` $180^{o}$ ( `2` góc kề bù )

$45^{o}$ `+` $\widehat{FCB}$`=` $180^{o}$

`=>` $\widehat{FCB}$ `=` $135^{o}$ `(1)`

Ta có: ( $\widehat{EAY}$ `=` $45^{o}$ vì $\widehat{EAY}$`=`$\widehat{BAD}$( `2` góc đối đỉnh ) 

$\widehat{BAX}$`+`$\widehat{XAE}$`+`$\widehat{EAY}$`=`$180^{o}$ ( `3` góc kề bù )

$90^{o}$ `+` $\widehat{XAE}$ `+` $45^{o}$ `=` $180^{o}$

`=>` $\widehat{XAE}$ `=` $45^{o}$

`=>` $\widehat{BAX}$`+`$\widehat{XAE}$`=` $\widehat{BAE}$ `=` $135^{o}$``(2)`

Từ `(1)(2)` 

`=>` $\widehat{BAE}$ `=` $\widehat{BCF}$ `=` $135^{o}$

Xét `ΔBAE` và `ΔBCF`, ta có:

`AE=BC`(gt)

`CF=AB`(gt)

$\widehat{BAE}$ `=` $\widehat{BCF}$ (cmt)

`=>` `ΔBAE` `=` `ΔBCF` (c.g.c)

`=>` `BE=BF` (đpcm)

`=>` Tam giác `BAE` cân tại `B`

`b)` Ta có:

$\widehat{EBA}$`=` $\widehat{BFC}$ (2 cạnh tương ứng )

$\widehat{ABF}$ `+` $\widehat{BFA}$ `=` $90^{o}$ ( tam giác `BAF` vuông tại `A`

mà $\widehat{ABF}$`=` $\widehat{ABC}$`+`$\widehat{CFB}$

`=>` $\widehat{ABC}$ `+` $\widehat{CFB}$ `+` $\widehat{BFA}$ `=` $90^{o}$

`=>` $\widehat{ABC}$ `+` $\widehat{CFB}$ `+` $\widehat{EBA}$ `=` $90^{o}$

mà  $\widehat{EBF}$ `=` $\widehat{EBA}$ `+` $\widehat{ABC}$ `+` $\widehat{CBF}$

`=>` $\widehat{EBF}$ `=` $90^{o}$ và `BE⊥BF`

 

Đáp án + Giải thích các bước giải :

`=>` `text{BE = FC}` (`2` cạnh tương ứng)

Ta có :

`\hat{\text{EBA}}=\hat{\text{BFC}}` (`2` góc tương ứng)

`\hat{\text{ABF}}+\hat{\text{BFA}}=90^\text{o}` `(\text{ΔABF ⊥ A})` mà `\hat{\text{ABF}}=\hat{\text{ABC}}+\hat{\text{CFB}}` 

`=>` `(\hat{\text{ABC}}+\hat{\text{CFB}})+\hat{text{BFA}}=90^\text{o}`

`=>` `\hat{\text{ABC}}+\hat{\text{CFB}}+\hat{text{EBA}}=90^\text{o}` mà `\hat{\text{EBF}}=\hat{\text{EBA}}+\hat{\text{ABC}}+\hat{\text{CBF}}`

`=>` `\hat{\text{EBF}}=90^\text{o}` `=>` `\text{BE ⊥ BF}`