a)
Xét $\Delta ABM$ và $\Delta ACN$, ta có:
$\widehat{BAC}$ là góc chung
$\widehat{ABM}=\widehat{ACN}\,\,\,\left( gt \right)$
$\to \Delta ABM\backsim\Delta ACN\,\,\,\left( g.g \right)$
b)
Vì $\Delta ABM\backsim\Delta ACN\,\,\,\left( cmt \right)$
$\to \dfrac{AB}{AC}=\dfrac{AM}{AN}\,\,\,\to \,\,\,\frac{AM}{AB}=\frac{AN}{AC}$
Xét $\Delta AMN$ và $\Delta ABC$, ta có:
$\widehat{BAC}$ là góc chung
$\dfrac{AM}{AB}=\dfrac{AN}{AC}\,\,\,\left( cmt \right)$
$\to \Delta AMN\backsim\Delta ABC\,\,\,\left( c.g.c \right)$
$\to \widehat{AMN}=\widehat{ABC}$ ( hai góc tương ứng )
c)
Xét $\Delta IBN$ và $\Delta ICM$, ta có:
$\widehat{IBN}=\widehat{ICM}\,\,\,\left( gt \right)$
$\widehat{BIN}=\widehat{CIM}$ ( hai góc đối đỉnh )
$\to \Delta IBN\backsim\Delta ICM\,\,\,\left( g.g \right)$
$\to \dfrac{IB}{IC}=\dfrac{IN}{IM}$
$\to IB.IM=IC.IN$
