a)

$AB=4cm\,\,;\,\,BC=20cm\,\,;\,\,CD=25cm$

$AD=8cm\,\,;\,\,BD=10cm$

Ta có:

+   $\dfrac{AB}{BD}=\dfrac{4}{10}=\dfrac{2}{5}$

+   $\dfrac{AD}{BC}=\dfrac{8}{20}=\dfrac{2}{5}$

+   $\dfrac{BD}{DC}=\dfrac{10}{25}=\dfrac{2}{5}$

$\to \dfrac{AB}{BD}=\dfrac{AD}{BC}=\dfrac{BD}{DC}=\dfrac{2}{5}$

$\to \Delta ABD\backsim\Delta BDC\left( c.c.c \right)$

b)

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

$\Rightarrow \widehat{ABD}=\widehat{BDC}$

$\Rightarrow \dfrac{1}{2}\widehat{ABD}=\dfrac{1}{2}\widehat{BDC}$

$\Rightarrow \widehat{MBD}=\widehat{BDN}$

Mà hai góc này nằm ở vị trí so le trong

Nên $BM//DN$

c)

$\Delta ABD$ có phân giác $BM\Rightarrow \dfrac{AM}{MD}=\dfrac{BA}{BD}$

$\Delta BDC$ có phân giác $DN\Rightarrow \dfrac{BN}{NC}=\dfrac{DB}{DC}$

Vì $\Delta ABD\backsim\Delta BDC\left( cmt \right)\Rightarrow \dfrac{BA}{BD}=\dfrac{DB}{DC}$

$\Rightarrow \dfrac{AM}{MD}=\dfrac{DN}{NC}$