Question

Answer 1

Ta có : \(OM=OA+AM=3+3=6\left(cm\right);ON=OB+BN=2+7=9\left(cm\right)\)

Xét ΔOBA và ΔOMN có :

\(\dfrac{OB}{OA}=\dfrac{2}{6}=\dfrac{1}{3}\)

\(\dfrac{OA}{AN}=\dfrac{3}{9}=\dfrac{1}{3}\)

\(\Rightarrow\dfrac{OB}{OM}=\dfrac{OA}{ON}\)

Mà \(\widehat{O}:chung\)

\(\Rightarrow\Delta OBA\sim\Delta OMN\left(c-g-c\right)\)

\(\Rightarrow\dfrac{OA}{ON}=\dfrac{OB}{OM}\)

Xét ΔOAN và ΔOBM có :

\(\dfrac{OA}{OB}=\dfrac{ON}{OM}\left(suy\cdot ra\cdot từ\cdot\dfrac{OA}{ON}=\dfrac{OB}{OM}\right)\)

Mà \(\widehat{O}:chung\)

\(\Rightarrow\Delta OAN\sim\Delta OBM\left(c-g-c\right)\)