góc IBC+góc ICB=1/2(góc ABC+góc ACB)
=1/2(180 độ-70 độ)=55 độ
=>góc BIC=125 độ
góc KBC+góc KCB
\(=\dfrac{180^0-\widehat{ABC}}{2}+\dfrac{180^0-\widehat{ACB}}{2}\)
\(=\dfrac{360^0-\left(180^0-70^0\right)}{2}=\dfrac{360^0-110^0}{2}=125^0\)
=>góc BKC=55 độ