a)

$M,P$ lần lượt là trung điểm $BC,NC$

$\Rightarrow MP$ là đường trung bình $\Delta NBC$

$\Rightarrow MP//BN$

$\Rightarrow MP//KN$

$\Rightarrow \Delta AKN\backsim\Delta AMP$

$\Rightarrow \dfrac{{{S}_{\Delta AKN}}}{{{S}_{\Delta AMP}}}={{\left( \dfrac{AK}{AM} \right)}^{2}}={{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}$

b)

Vì $\Delta AKN\backsim\Delta AMP\left( cmt \right)$

$\Rightarrow \dfrac{AN}{AP}=\dfrac{AK}{AM}=\dfrac{1}{3}$

Mà do $P$ là trung điểm $NC$

Nên sẽ tính được: $\dfrac{AP}{AC}=\dfrac{3}{5}$

Kẻ $MF\bot AC$

Có:      ${{S}_{\Delta AMP}}=\dfrac{1}{2}MF.AP$

            ${{S}_{\Delta AMC}}=\dfrac{1}{2}MF.AC$

$\Rightarrow \dfrac{{{S}_{\Delta AMP}}}{{{S}_{\Delta AMC}}}=\dfrac{AP}{AC}=\dfrac{3}{5}$

Kẻ $AG\bot BC$

Có:      ${{S}_{\Delta AMC}}=\dfrac{1}{2}AG.MC$

            ${{S}_{\Delta ABC}}=\dfrac{1}{2}AG.BC$

$\Rightarrow \dfrac{{{S}_{\Delta AMC}}}{{{S}_{\Delta ABC}}}=\dfrac{MC}{BC}=\dfrac{1}{2}$

Tổng hợp lại, ta có: ${{S}_{\Delta AKN}}=\dfrac{1}{9}{{S}_{\Delta AMP}}$ ; ${{S}_{\Delta AMP}}=\dfrac{3}{5}{{S}_{\Delta AMC}}$ ; ${{S}_{\Delta AMC}}=\dfrac{1}{2}{{S}_{\Delta ABC}}$

$\Rightarrow {{S}_{\Delta AKN}}=\dfrac{1}{9}\cdot \dfrac{3}{5}\cdot \dfrac{1}{2}{{S}_{\Delta ABC}}=\dfrac{1}{30}{{S}_{\Delta ABC}}$

c)

Vẽ $BE//IJ$,$CH//IJ$ $\left( E,H\in AM \right)$

$\Rightarrow BE//CH$

Xét $\Delta EBM$ và $\Delta HCM$, ta có:

$MB=MC$ ($M$ trung điểm $BC$)

$\widehat{EMB}=\widehat{HMC}$ (đối đỉnh)

$\widehat{EBM}=\widehat{HCM}$ (so le trong)

$\Rightarrow \Delta EBM=\Delta HCM\left( g.c.g \right)$

$\Rightarrow EM=HM$

$\Rightarrow EM-HM=0$

Theo định lý Ta-let, ta có:

$\dfrac{AB}{AI}=\dfrac{AE}{AK}$;$\dfrac{AC}{AJ}=\dfrac{AH}{AK}$

$\Rightarrow \dfrac{AB}{AI}+\dfrac{AC}{AJ}=\dfrac{AE}{AK}+\dfrac{AH}{AK}$

$\Rightarrow \dfrac{AB}{AI}+\dfrac{AC}{AJ}=\dfrac{1}{AK}\left( AE+AH \right)$

$\Rightarrow \dfrac{AB}{AI}+\dfrac{AC}{AJ}=\dfrac{1}{AK}\left( AM+EM+AM-HM \right)$

$\Rightarrow \dfrac{AB}{AI}+\dfrac{AC}{AJ}=\dfrac{1}{AK}\left( 2AM \right)$

$\Rightarrow \dfrac{AB}{AI}+\dfrac{AC}{AJ}=\dfrac{6AK}{AK}=6$