Giải thích các bước giải:

 Kẻ $IG\perp BC, ID\perp AB, IF\perp AC$

        $IG\cap MN=E$

Vì $MN//BC$

$\to IE\perp MN$

Ta có: 

$MD, ME$ là tiếp tuyến của $(I)\to MD=ME$

$NE, NF$ là tiếp tuyến của $(I)\to NE=NF$

$\to MD+NF=ME+EN=MN=2$

Ta có:

$MN//BC\to \widehat{AMN}=\widehat{ABC},\widehat{ANM}=\widehat{ACB}$

$\to \Delta AMN\sim\Delta ABC(g.g)$

$\to \dfrac{P_{AMN}}{P_{ABC}}=\dfrac{MN}{CB}$

$\to \dfrac{AM+AN+MN}{16}=\dfrac{2}{CB}$

$\to \dfrac{AM+AN+ME+EN}{16}=\dfrac{2}{CB}$

$\to \dfrac{AM+AN+MD+NF}{16}=\dfrac{2}{CB}$

$\to \dfrac{AD+AF}{16}=\dfrac{2}{CB}$

$\to \dfrac{(AB-BD)+(AC-CF)}{16}=\dfrac{2}{CB}$

$\to \dfrac{AB+AC-(BD+CF)}{16}=\dfrac{2}{CB}$

$\to \dfrac{AB+AC-(BG+GC)}{16}=\dfrac{2}{CB}$

$\to \dfrac{AB+AC-BC}{16}=\dfrac{2}{CB}$

$\to \dfrac{AB+AC+BC-2BC}{16}=\dfrac{2}{CB}$

$\to \dfrac{16-2BC}{16}=\dfrac{2}{CB}$

$\to  16BC-2BC^2=22$

$\to 8BC-BC^2=16$

$\to BC^2-8BC+16=0$

$\to BC^2-8BC+16=0$

$\to (BC-4)^2=0$

$\to BC-4=0$

$\to BC=4$