a) Có:$ \widehat{AMK}=\widehat{BMK}$ ( đều là góc nội tiếp chắn 2 cung bằng nhau )
Ta lại có:$\widehat{AMK}+\widehat{BMK}+\widehat{DMB}=180$*(1)
$\widehat{DMB}+\widehat{BDM}+\widehat{DBM}=180$*(2)
Từ $(1) $ và $(2) $=> $\widehat{AMK}+\widehat{BMK}+\widehat{DMB}=\widehat{DMB}+\widehat{BDM}+\widehat{DBM}$
<=>$ \widehat{AMK}+\widehat{BMK}=\widehat{BDM}+\widehat{DBM}$
<=>$2\widehat{AMK}=2\widehat{BDM}$ ( vì góc BDM = góc DBM )
<=>$\widehat{AMK}=\widehat{BDM}$
Có:$ \widehat{MAB}+\widehat{AMK}+\widehat{AQM}=180$*
$ \widehat{DAB}+\widehat{BDA}+\widehat{ABD}=180$*
VÌ $ A,M,D$ thẳng hàng:=>$\widehat{AQM}=\widehat{ABD}$
Ta có: $\widehat{AQM}=\widehat{KQB}$ (đđ)
=> $\widehat{KQB}=\widehat{ABD}$
b) Dễ thấy $BM=MD$
$ AM>CM$
Xét $\triangle DMC$ có:
$DM+DM>CD$
=> $MA+MD>CD$
<=>$MA+MB>CD$