Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta CMD$  có:

$MA=MC$

$\widehat{AMB}=\widehat{CMD}$

$MB=MD$

$\to\Delta AMB=\Delta CMD(c.g.c)$

b.Từ câu a $\to AB=CD, \widehat{MAB}=\widehat{MCD}\to AB//CD\to\widehat{BAC}=\widehat{ACD}$

Xét $\Delta ABC,\Delta ACD$ có:

Chung $AC$

$\widehat{BAC}=\widehat{ACD}$

$AB=CD$

$\to\Delta ABC=\Delta CDA(c.g.c)$

c.Từ câu b $\to AD=BC,\widehat{CAD}=\widehat{ACB}\to AD//BC$

d.Xét $\Delta NAK,\Delta NCB$ có:

$NA=NB$

$\widehat{ANK}=\widehat{BNC}$

$NK=NC$

$\to\Delta ANK=\Delta BNC(c.g.c)$

$\to\widehat{NAK}=\widehat{NBC}\to AK//BC$

Mà $AD//BC$
$\to K, A,D $ thẳng hàng

e.Từ câu  $\to AB=CD, \widehat{ABC}=\widehat{ADC}\to \widehat{ABF}=\widehat{CDE}$

Xét $\Delta ABF,\Delta CDE$ có:

$\hat F=\hat E(=90^o)$

$AB=CD$

$\widehat{ABF}=\widehat{EDC}$

$\to \Delta ABF=\Delta CDE$(cạnh huyền-góc nhọn)

$\to BF=DE$