Có
$\widehat{DBC}=90^c - 60^o=30^o$
Suy ra $\widehat{CBK}=150 ^o$
Tam giác BKC cân ở B nên $\widehat{BKC}=\widehat{BCK}=\dfrac{180^o-\widehat{KBC}}{2}=15^o$
Tương tự có $\widehat{AKB}=15^o$
-> $\widehat{AKC}=30^o$
-> $\widehat{AKC} + \widehat{ABC} = 30 ^o + 60^o = 90^o$