Xét tg ABC có
\(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-68^o=112^o\)
\(\widehat{KBC}=\dfrac{\widehat{ABx}-\widehat{B}}{2}=\dfrac{180^o-\widehat{B}}{2}=90^o-\dfrac{\widehat{B}}{2}\)
\(\widehat{KCB}=\dfrac{\widehat{ACy}-\widehat{C}}{2}=\dfrac{180^o-\widehat{C}}{2}=90^o-\dfrac{\widehat{C}}{2}\)
Xét tg KBC có
\(\widehat{BKC}=180^o-\left(\widehat{KBC}+\widehat{KCB}\right)=\)
\(=180^o-\left(90^o-\dfrac{\widehat{B}}{2}+90^o-\dfrac{\widehat{C}}{2}\right)=\dfrac{\widehat{B}+\widehat{C}}{2}=\dfrac{112^o}{2}=56^o\)