Giải thích các bước giải:
a.Ta có $MA=MD, MB=MC, \widehat{AMB}=\widehat{CMD}\to\Delta ABM=\Delta DMC(c.g.c)\to AB=CD$
b.Ta có $MA=MD,\widehat{AHM}=\widehat{MKD}=90^o,\widehat{AMH}=\widehat{DMK}$
$\to\Delta AHM=\Delta DKM(g.c.g)\to AH=DK$
c.Từ câu b $\to AH=DK\to \dfrac 12 AH=\dfrac 12 DK\to AE=DF$
Mà $\widehat{EAM}=\widehat{MDF}, AM=MD\to \Delta AEM=\Delta DFM(c.g.c)$
$\to\widehat{AME}=\widehat{DMF}$
$\to\widehat{EMF}=\widehat{AME}+\widehat{AMF}=\widehat{DMF}+\widehat{AMF}=\widehat{AMD}=180^o$
$\to E, M,F $ thẳng hàng