Giải thích các bước giải:

Ta có:

$AE=\dfrac12BE$

$\to AE+BE=\dfrac32BE$

$\to AB=\dfrac32BE$

$\to \dfrac{BE}{BA}=\dfrac23$

Kẻ $EF//AC, F\in BD$

$\to \dfrac{EF}{AD}=\dfrac{BE}{BA}=\dfrac23$

$\to EF=\dfrac23AD=\dfrac23\cdot \dfrac13CD=\dfrac29CD$

$\to \dfrac{EF}{CD}=\dfrac29$

Lại có: $EF//CD$

$\to \dfrac{EK}{KC}=\dfrac{EF}{CD}=\dfrac29$