a: \(\widehat{ABC}=180^0-70^0-30^0=80^0\)
\(\widehat{ABD}=\dfrac{80^0}{2}=40^0\)
\(\widehat{ADB}=180^0-70^0-40^0=70^0\)
b: \(\widehat{IBC}+\widehat{ICB}=\dfrac{\widehat{ABC}}{2}+\dfrac{\widehat{ACB}}{2}=\dfrac{1}{2}\left(30^0+80^0\right)=55^0\)
=>\(\widehat{BIC}=125^0\)
\(\widehat{CID}=180^0-125^0=55^0\)