Từ C kẻ CK( \(K\in DN\)) sao cho OM//CK suy ra OD=OK( vì M là tđ DC)
Ta có \(\frac{OD}{ON}=\frac{2}{3}\Leftrightarrow\frac{OK}{ON}=\frac{2}{3}\left(1\right)\)
Nhân (1) với 1 được: \(\frac{OK}{ON}.\frac{OK}{OK}=\frac{OK}{ON}=\frac{2}{3}\)
vậy ta có: \(\frac{ON}{OK}=\frac{3}{2}\Leftrightarrow\frac{ON}{OK}-1=\frac{1}{2}\Leftrightarrow\frac{NK}{OK}=\frac{1}{2}\Leftrightarrow\frac{NK}{OD}=\frac{1}{2}\)(2)
Ta lại có: \(KC=2OM=2.\frac{1}{4}OA=\frac{1}{2}OA\Leftrightarrow\frac{KC}{OA}=\frac{1}{2}\left(3\right)\)
ta có: AM//CK nên \(\widehat{AON}=\widehat{OKC}\Leftrightarrow\widehat{AOD}=\widehat{CKN}\left(4\right)\)
(2),(3) và (4) suy ra \(\Delta AOD\sim\Delta CKN\Rightarrow\widehat{ADO}=\widehat{CNK}\Rightarrow\)AD//NC( so le trong)