Giải thích các bước giải:
a.Xét $\Delta AMC, \Delta DMB$ có:
$MA=MD$
$\widehat{AMC}=\widehat{BMD}$(đối đỉnh)
$MC=MB$ vì $M$ là trung điểm $BC$
$\to \Delta AMC=\Delta DMB(c.g.c)$
$\to AC=BD, \widehat{MAC}=\widehat{MDB}\to AC//BD$
b.Xét $\Delta AIM,\Delta DMK$ có:
$MA=MD$
$\widehat{MAI}=\widehat{MDK}$ vì $AC//DB$
$AI=DK$
$\to \Delta AMI=\Delta DMK(c.g.c)$
$\to\widehat{AMI}=\widehat{KMD}$
$\to I,M,K$ thẳng hàng
c.Xét $\Delta ACE, \Delta ABF$ có:
$AE=AB$
$\widehat{EAC}=90^o+\widehat{BAC}=\widehat{BAF}$
$AC=AF$
$\to \Delta AEC=\Delta ABF(c.g.c)$
$\to \widehat{AEC}=\widehat{ABF}$
Gọi $AB\cap CE=G, BF\cap CE=H$
$\to \widehat{GEA}=\widehat{GBH}$
$\to \widehat{GHB}=180^o-\widehat{GBH}-\widehat{HGB}=180^o-\widehat{AEG}-\widehat{EGA}=90^o$
$\to BF\perp CE$