Giải thích các bước giải:
a.Xét $\Delta AMB, \Delta KMC$ có:
$MA=MK$
$\widehat{AMB}=\widehat{CMK}$
$MB=MC$
$\to \Delta AMB=\Delta KMC(c.g.c)$
$\to \widehat{KCM}=\widehat{ABM}\to AB//CK$
$\to \widehat{ACK}=180^o-\widehat{BAC}=70^o$
b.Từ câu a $\to CK=AB=AD$
Xét $\Delta ADE, \Delta CKA$ có:
$AE=AK$
$\widehat{DAE}=360^o-\widehat{DAB}-\widehat{EAC}-\widehat{BAC}=360^o-90^o-90^o-110^o=70^o=\widehat{ACK}$
$AD=CK$
$\to \Delta ADE=\Delta CKA(c.g.c)$
c.Gọi $MA\cap DE=H$
Từ câu b
$\to \widehat{KAC}=\widehat{AED}$
$\to \widehat{KAC}=\widehat{HEA}$
$\to \widehat{DEA}+\widehat{DAE}=\widehat{KAC}+\widehat{HAE}=180^o-\widehat{EAC}=90^o$
$\to \Delta HAE$ vuông tại $H$
$\to AH\perp HE\to MA\perp DE$