Giải thích các bước giải:
Ta có:
$\widehat{FBE}+\widehat{BEF}+\widehat{EFB}=180^o$
$\widehat{FDE}+\widehat{DEF}+\widehat{DFE}=180^o$
$\to (\widehat{FBE}+\widehat{BEF}+\widehat{EFB})+(\widehat{FDE}+\widehat{DEF}+\widehat{DFE})=360^o$
$\to(\widehat{FBE}+\widehat{FDE})+(\widehat{DEF}+\widehat{DFE})+(\widehat{BEF}+\widehat{BFE})=360^o$
Mà $\Diamond ABCD$ có các góc đối bù nhau
$\to\widehat{ABC}+\widehat{ADC}=180^o$
$\to\widehat{FBE}+\widehat{EDF}=180^o$
$\to 180^o+(\widehat{DEF}+\widehat{DFE})+(\widehat{BEF}+\widehat{BFE})=360^o$
$\to (\widehat{DEF}+\widehat{DFE})+(\widehat{BEF}+\widehat{BFE})=180^o$
$\to (\widehat{DEF}+\widehat{DFE})+(\widehat{BFC}+\widehat{DFE}+\widehat{BEA}+\widehat{DEF})=180^o$
$\to 2(\widehat{DEF}+\widehat{DFE})+(\widehat{BFC}+\widehat{BEA})=180^o$
$\to 2(\widehat{DEF}+\widehat{DFE})+(2\widehat{MFC}+2\widehat{MEA})=180^o$ vì $FM,EM$ là phân giác $\widehat{BFC},\widehat{BEA}$
$\to \widehat{DEF}+\widehat{DFE}+\widehat{MFC}+\widehat{MEA}=90^o$
$\to\widehat{MFE}+\widehat{MEF}=90^o$
$\to \Delta MEF$ vuông tại $M$
$\to FM\perp EM$
