a)

$\dfrac{AM}{MB}=\dfrac{3}{2}\,\,\,;\,\,\dfrac{AN}{NC}=\dfrac{7,5}{5}=\dfrac{3}{2}$

$\to \dfrac{AM}{MB}=\dfrac{AN}{NC}$

$\to MN\,\,||\,\,BC$

b)

$\Delta ABI$:

$MK\,\,||\,\,BI$

$\to \dfrac{AK}{AI}=\dfrac{MK}{BI}$

$\Delta ACI$:

$NK\,\,||\,\,CI$

$\to \dfrac{AK}{AI}=\dfrac{NK}{CI}$

$\to \dfrac{MK}{BI}=\dfrac{NK}{CI}$

Mà $BI=CI$ ( vì $I$ là trung điểm $BC$ )

Nên $MK=NK$

$\to K$ là trung điểm $MN$

c)

$\Delta OMN$ có $MN\,\,||\,\,BC$

$\to \dfrac{MO}{CO}=\dfrac{MN}{BC}=\dfrac{\dfrac{1}{2}MN}{\dfrac{1}{2}BC}=\dfrac{MK}{CI}$

Xét $\Delta OMK$ và $\Delta OCI$, ta có:

$\widehat{OMK}=\widehat{OCI}$ ( $MN\,\,||\,\,BC$, hai góc so le trong )

$\dfrac{MO}{CO}=\dfrac{MK}{CI}\,\,\,\left( \,cmt\, \right)$

$\to \Delta OMK\sim\Delta OCI\,\,\,\left( \,c\,.\,g\,.\,c\, \right)$

$\to \widehat{MOK}=\widehat{COI}$ ( hai góc tương ứng )

Mà $\widehat{MOK}+\widehat{COK}=180{}^\circ $ ( hai góc kề bù )

Nên $\widehat{COI}+\widehat{COK}=180{}^\circ $

$\to K,O,I$ thẳng hàng

$\to A,O,I$ thẳng hàng