c)

Vì $\Delta ABD\backsim\Delta ACE\left( cmt \right)$

$\Rightarrow \dfrac{AB}{AC}=\dfrac{AD}{AE}$

$\Rightarrow \dfrac{AD}{AB}=\dfrac{AE}{AC}$

Xét $\Delta ADE$ và $\Delta ABC$, ta có

+ $\dfrac{AD}{AB}=\dfrac{AE}{AC}\left( cmt \right)$

+ $\widehat{BAC}$ là góc chung

Nên $\Delta ADE\backsim\Delta ABC\left( c.g.c \right)$

$\Rightarrow \widehat{ADE}=\widehat{ABC}$

 

d)

Vì $\Delta ABD\backsim\Delta ACE\left( cmt \right)$

$\Rightarrow \dfrac{AB}{AC}=\dfrac{AD}{AE}\Rightarrow AE.AB=AD.AC\,\,\,\left( 1 \right)$

 

Xét $\Delta AEN$ và $\Delta ANB$, ta có:

$\widehat{NAE}$ là góc chung

$\widehat{AEN}=\widehat{ANB}=90{}^\circ $

Nên $\Delta AEN\backsim\Delta ANB\left( g.g \right)$

Do đó $\dfrac{AE}{AN}=\dfrac{AN}{AB}\Rightarrow A{{N}^{2}}=AE.AB\,\,\,\left( 2 \right)$

 

Xét $\Delta ADM$ và $\Delta AMC$, ta có:

$\widehat{DAM}$ là góc chung

$\widehat{ADM}=\widehat{AMC}=90{}^\circ $

Nên $\Delta ADM\backsim\Delta AMC\left( g.g \right)$

Do đó $\dfrac{AD}{AM}=\dfrac{AM}{AC}\Rightarrow A{{M}^{2}}=AD.AC\,\,\,\left( 3 \right)$

 

Từ $\left( 1 \right),\left( 2 \right),\left( 3 \right)\Rightarrow A{{M}^{2}}=A{{N}^{2}}\Rightarrow AM=AN$