a)

$IJ$ là đường trung bình $\Delta DAB$

$\to IJ\,//\,AB$

$\to \widehat{DIJ}=\widehat{DAB}\,\,\,\,\,\left( 1 \right)$

$IK$ là đường trung bình $\Delta DAC$

$\to IK\,//\,AC$

$\to \widehat{DIK}=\widehat{DAC}\,\,\,\,\,\left( 2 \right)$

$\left( 1 \right)+\left( 2 \right)\Leftrightarrow \widehat{DIJ}+\widehat{DIK}=\widehat{DAB}+\widehat{DAC}$

$\left( 1 \right)+\left( 2 \right)\Leftrightarrow \widehat{JIK}=\widehat{BAC}$

Chứng minh tương tự, $\widehat{IJK}=\widehat{ABC}$

$\to \Delta IJK\backsim\Delta ABC\,\,\,\left( g.g \right)$

b)

Vì $\Delta IJK\backsim\Delta ABC\,\,\,\left( cmt \right)$

$\to \dfrac{{{S}_{\Delta IJK}}}{{{S}_{\Delta ABC}}}={{\left( \dfrac{IJ}{AB} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}$

$\to {{S}_{\Delta IJK}}=\dfrac{{{S}_{\Delta ABC}}}{4}=\dfrac{18}{4}=4,5\,\,\,\left( c{{m}^{2}} \right)$

c)

$\Delta APM$ có $IJ\,//\,AP\,\,\Rightarrow \dfrac{AI}{AM}=\dfrac{AP}{IJ}$

$\Delta AQM$ có $IK\,//\,AM\,\,\Rightarrow \dfrac{AI}{AM}=\dfrac{AQ}{IK}$

$\Rightarrow \dfrac{AP}{IJ}=\dfrac{AQ}{IK}$

$\Rightarrow \Delta APQ\backsim\Delta IJK\,\,\,\left( c.g.c \right)$

$\Rightarrow \Delta APQ\backsim\Delta ABC$

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

$\Rightarrow BC\,//\,PQ$

$\Rightarrow JK//PQ$