Đáp án:
$\begin{array}{l}
\widehat {BIC} = {180^0} - \widehat {CID} = {180^0} - {60^0} = {120^0}\\
Do:\widehat {ABI} = \widehat {IBC} = \frac{1}{2}\widehat {ABC}\\
\widehat {ACI} = \widehat {ICB} = \frac{1}{2}\widehat {ACB}\\
\Rightarrow \widehat {IBC} + \widehat {ICB} = \frac{1}{2}\left( {\widehat {ABC} + \widehat {ACB}} \right)\\
\Rightarrow {180^0} - \widehat {BIC} = \frac{1}{2}\left( {\widehat {ABC} + \widehat {ACB}} \right)\\
\Rightarrow \frac{1}{2}\left( {\widehat {ABC} + \widehat {ACB}} \right) = {60^0}\\
\Rightarrow \widehat {ABC} + \widehat {ACB} = {120^0}\\
\Rightarrow \widehat {BAC} = {180^0} - {120^0} = {60^0}
\end{array}$
$\begin{array}{l}
\widehat {BEC} + \widehat {BDC}\\
= \left( {{{180}^0} - \widehat {ABC} - \widehat {ECB}} \right) + \left( {{{180}^0} - \widehat {ACB} - \widehat {DBC}} \right)\\
= {360^0} - \widehat {ABC} - \widehat {ACB} - \frac{1}{2}\widehat {ACB} - \frac{1}{2}\widehat {ABC}\\
= {360^0} - {120^0} - \frac{1}{2}.\left( {\widehat {ABC} + \widehat {ACB}} \right)\\
= {360^0} - {120^0} - \frac{1}{2}{.120^0}\\
= {180^0}
\end{array}$
Vậy góc BEC và BDC bù nhau
