Xét tam giác BIC có:

\(\widehat{IBC}={{\widehat{I}}_{1}}-\widehat{BCI}\)

Xét tam giác DIC có:

\(\widehat{IDC}={{\widehat{I}}_{2}}-\widehat{ICD}\)

Nên \(\widehat{BCI}+\widehat{ICD}=({{\widehat{I}}_{1}}+{{\widehat{I}}_{2}})-({{\widehat{C}}_{1}}+{{\widehat{C}}_{2}})=\widehat{BID}-\widehat{C}\)

Tứ giác ABID có:

\(\widehat{ABI}+\widehat{ADI}={{360}^{o}}-\widehat{A}-\widehat{BID}\) (Do \(\widehat{ABI}=\widehat{IBC}\) và \(\widehat{ADI}=\widehat{IDC}\))

Nên \(\widehat{IBC}+\widehat{IDC}=\widehat{ABI}+\widehat{ADI}\)

\(\Rightarrow \widehat{BID}-\widehat{C}={{360}^{o}}-\widehat{A}-\widehat{BID}\)

\(\Rightarrow 2\widehat{BID}={{360}^{o}}-(\widehat{A}-\widehat{C})={{360}^{o}}-{{60}^{o}}={{300}^{o}} \)

Suy ra \(\widehat{BID}={{150}^{o}}\)

Vậy chọn A.