Xét $\Delta CNB$ và $\Delta CMA$, ta có:
$\widehat{ACB}$ là góc chung
$\widehat{CNB}=\widehat{CMA}=90{}^\circ $
$\to \Delta CNB\backsim\Delta CMA\,\,\,\left( g.g \right)$
$\to \dfrac{CN}{CM}=\dfrac{CB}{CA}$
$\to \dfrac{CA}{CM}=\dfrac{CB}{CN}$
Xét $\Delta ABC$ và $\Delta MNC$, ta có:
$\widehat{ACB}$ là góc chung
$\dfrac{CA}{CM}=\dfrac{CB}{CN}\,\,\,\left( cmt \right)$
$\to \Delta ABC\backsim\Delta MNC\,\,\,\left( c.g.c \right)$