dat BN=y va CM=x va goi giao diem CM ;BN là O
\(\Rightarrow OM\)\(=\dfrac{1}{3}x;ON=\dfrac{1}{3}y;OB=\dfrac{2}{3}y;OC=\dfrac{2}{3}x\)
\(\Delta\) BOM cho:\(\left(\dfrac{1}{3}x\right)^2+\left(\dfrac{2}{3}y\right)^2=\left(\dfrac{19}{2}\right)^2\left(1\right)\) \(\Delta\)CON cho:\(\left(\dfrac{2}{3}x\right)^2+\left(\dfrac{1}{3}y\right)^2=11^2\left(2\right)\)
từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x^2=144\\y^2=\dfrac{945}{4}\end{matrix}\right.\left(3\right)\)
\(\Delta\)MON cho : \(MN^2=OM^2+ON^2=\dfrac{1}{9}\left(144+\dfrac{945}{4}\right)\Rightarrow MN=\dfrac{13}{2}\) ( do (3) )
lại có MN là đường trung bình tam giác ABC \(\Rightarrow BC=2MN=13\)
