Xét \(\bigtriangleup{ABE}\) vuông tại A có AG \(\perp BE = \) {G}
Áp dụng hệ thức \(c^2 = a . c'\)
\(\Leftrightarrow\) \(AB^2 = BE . BG\)
Vì AD \(\cap BE \) = {G}
\(\Rightarrow\) BG = \(\dfrac{2}{3}\) BE ( tính chất)
\(\Rightarrow\) AB = BE . \(\dfrac{2}{3}\) BE
\(\Leftrightarrow\) \(\sqrt{6}^2\) = \(BE ^2 . \dfrac{2}{3} \)
\(\Leftrightarrow\) 6 = \(\dfrac{2}{3}\) . \(BE^2 \)
\(\Leftrightarrow\) \(BE^2=9 = (\pm 3)^2 \)
Vì\( BE >0 \)
\(\Rightarrow\) \(BE= 3\)
Áp dụng định lý Py-ta-go , có:
\(BE^2 = AB^2+AE^2 \)
\(\Leftrightarrow\) \(3^2 = \sqrt{6^2} + AE ^2\)
\(\Leftrightarrow\) \(9=6+AE^2\)
\(\Leftrightarrow\) \(AE^2 = 3\)
\(\Rightarrow\)\(AE = \sqrt{3}\)
Ta có : AE . EC = AC
\(\Leftrightarrow\) \(\sqrt{3} . \sqrt{3} = AC \)
\(\Leftrightarrow\) AC = \(2\sqrt{3}\)
Áp dụng định lý Py-ta-go , có :
\(BC^2 =AB^2+AC^2 \)
\(\Leftrightarrow\) \(BC^2 = \sqrt{6^2} +(2\sqrt{3})^2\)
\(\Leftrightarrow\) \(BC^2 = 6+12 \)
\(\Leftrightarrow\) 
\(BC^2 = 18\)
\(\Rightarrow\) \(BC = \sqrt{18} = 2\sqrt{3}\)
Ta có : SinC = \(\dfrac{AB}{AC}\) = \(\dfrac{\sqrt{6}}{3\sqrt{2}}\) = \(\dfrac{\sqrt{3}}{3}\)
\(\Rightarrow\)\(\widehat{C} \) \(\approx 35 ^0\)
\(\Rightarrow\) \(\widehat{B} \approx 90^0 - 35^0=55^0\)
