Giải thích các bước giải:
Ta có $\dfrac{AE}{AD}=\dfrac13$
$\to \dfrac{AE}{AD-AE}=\dfrac{1}{3-1}$
$\to\dfrac{EA}{ED}=\dfrac12$
Kẻ $AF//BC$
$\to \dfrac{AF}{BD}=\dfrac{EA}{ED}=\dfrac12$
$\to AF=\dfrac12BD$
Mà $\dfrac{BD}{BC}=\dfrac34\to BD=\dfrac34BC$
$\to AF=\dfrac12\cdot \dfrac34BC$
$\to AF=\dfrac32BC$
$\to \dfrac{AF}{BC}=\dfrac32$
Lại có $AF//BC$
$\to \dfrac{KA}{KC}=\dfrac{AF}{BC}=\dfrac32$