a: Xét ΔDAC có MI//AC
nên \(\frac{DM}{DA}=\frac{DI}{DC}\)
mà \(\frac{DM}{DA}=\frac{BN}{BC}\)
nên \(\frac{DI}{DC}=\frac{BN}{BC}\)
b: Ta có: \(\frac{DI}{DC}=\frac{BN}{BC}\)
=>\(1-\frac{DI}{DC}=1-\frac{BN}{BC}\)
=>\(\frac{CI}{CD}=\frac{CN}{CB}\)
Xét ΔCBD có \(\frac{CN}{CB}=\frac{CI}{CD}\)
nên NI//BD



