$\widehat{BAQ}$= $\widehat{NAC}$ (=$\widehat{BAC}$+ $90^{0}$)=> $\Delta$BAQ= $\Delta$NAC=> BQ= NC và $\widehat{ABQ}$= $\widehat{ANC}$Xét tứ giác MBON có: $\widehat{BON}$+ $\widehat{ONM}$+ $\widehat{NMB}$+ $\widehat{MBO}$= $\widehat{BON}$+ $\widehat{ONM}$+ $90^{0}$+ $90^{0}$+ $\widehat{ABQ}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+ $\widehat{ONM}$+ $\widehat{ANC}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+$90^{0}$= $360^{0}$ => $\widehat{BON}$= $90^{0}$ Vậy BQ vuôgn góc với NC.
$\widehat{BAQ}$= $\widehat{NAC}$ (=$\widehat{BAC}$+ $90^{0}$)=> $\Delta$BAQ= $\Delta$NAC=> BQ= NC và $\widehat{ABQ}$= $\widehat{ANC}$Xét tứ giác MBON có: $\widehat{BON}$+ $\widehat{ONM}$+ $\widehat{NMB}$+ $\widehat{MBO}$= $\widehat{BON}$+ $\widehat{ONM}$+ $90^{0}$+ $90^{0}$+ $\widehat{ABQ}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+ $\widehat{ONM}$+ $\widehat{ANC}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+$90^{0}$= $360^{0}$ => $\widehat{BON}$= $90^{0}$ Vậy BQ vuôgn góc với NC.
$\widehat{BAQ}$= $\widehat{NAC}$ (=$\widehat{BAC}$+ $90^{0}$)=> $\Delta$BAQ= $\Delta$NAC=> BQ= NC và $\widehat{ABQ}$= $\widehat{ANC}$Xét tứ giác MBON có: $\widehat{BON}$+ $\widehat{ONM}$+ $\widehat{NMB}$+ $\widehat{MBO}$= $\widehat{BON}$+ $\widehat{ONM}$+ $90^{0}$+ $90^{0}$+ $\widehat{ABQ}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+ $\widehat{ONM}$+ $\widehat{ANC}$= $\widehat{BON}$+ $90^{0}$+ $90^{0}$+$90^{0}$= $360^{0}$ => $\widehat{BON}$= $90^{0}$ Vậy BQ vuôgn góc với NC.