Đáp án:
$\triangle BDC$ có $ON//CD$
$\to \dfrac{OD}{BD}=\dfrac{CN}{BC}$
$\triangle ABD$ có $OM//AB$
$\to \dfrac{AB}{OM}=\dfrac{BD}{OD}$
$\triangle ABC$ có $ON//AB$
$\to \dfrac{BC}{NC}=\dfrac{AB}{ON}$
$\to \dfrac{OM}{AB}=\dfrac{ON}{AB}$
$\to OM=ON(dpcm)$