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