Kéo dài BC và O'O cắt nhau tại D
Do OB song song O'C (cùng vuông góc BD), áp dụng định lý Thales:
\(\dfrac{DO}{DO'}=\dfrac{DB}{DC}=\dfrac{OB}{O'C}=\dfrac{2}{6}\) (1)
\(\Rightarrow\dfrac{DO}{DO+OO'}=\dfrac{1}{3}\)
\(\Rightarrow3DO=DO+8\)
\(\Rightarrow DO=4\)
\(\Rightarrow DB=\sqrt{DO^2-OB^2}=\sqrt{4^2-2^2}=2\sqrt{3}\)
\(\left(1\right)\Rightarrow\dfrac{DB}{DB+BC}=\dfrac{1}{3}\Rightarrow BC=2DB=4\sqrt{3}\)
Trong tam giác vuông DCO':
\(cos\widehat{O'}=\dfrac{O'C}{DO'}=\dfrac{6}{4+8}=\dfrac{1}{2}\Rightarrow\widehat{O'}=60^0\)
\(\Rightarrow\widehat{BOA}=180^0-\widehat{O'}=120^0\)
\(\Rightarrow\) Chu vi\(=BC+l_{\stackrel\frown{AB}}+l_{\stackrel\frown{AC}}=4\sqrt{3}+2.2\pi.\dfrac{120}{360}+6.2\pi.\dfrac{60}{360}=4\sqrt{3}+\dfrac{10\pi}{3}\left(cm\right)\)
