a)Gọi \(\widehat{CBx}\) và \(\widehat{BCy}\) lần lượt là 2 góc ngoài đỉnh B và C
Ta có:
\(\widehat{ABC}+\widehat{ACB}=180^0-\widehat{BAC}=180^0-50^0=130^0\)
\(\Rightarrow\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=\frac{1}{2}.130^0=65^0\)
\(\Rightarrow180^0-\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=180^0-65^0=115^0\)
\(\Rightarrow`180^0-\left(\widehat{IBC} +\widehat{ICB}\right)=115^0\)
\(\Rightarrow\widehat{BIC}=115^0\)
Tương tự, ta cũng có: \(\widehat{BKC}=180^0-\left(\widehat{KBC}+\widehat{KCB}\right)=180^0-\frac{1}{2}\left(\widehat{CBx}+\widehat{BCy}\right)=180^0-\frac{1}{2}\left[\left(180^0-\widehat{ABC}\right)+\left(180^0-\widehat{ACB}\right)\right]\)\(\Rightarrow\widehat{BKC}=180^0-\frac{1}{2}\left(360^0-130^0\right)=180^0-\frac{1}{2}.230=180^0-115=65^0\)b) Ta có:
\(\widehat{BDK}=180^0-\left(\widehat{DBK}+\widehat{DKB}\right)=180^0-\left(90^0+65^0\right)=180^0-155^0=25^0\)Hay \(\widehat{BDC}=25^0\)
