Đáp án: $a.EK=\dfrac{12}{5}$
Giải thích các bước giải:
a.Ta có $AB//CG\to\dfrac{AB}{DG}=\dfrac{AE}{EG}=\dfrac{BE}{ED}=\dfrac{3}{5}$
$\to AB=\dfrac35DG, EG=\dfrac53AE=\dfrac{20}{3}\to AG=AE+EG=\dfrac{32}{3}$
Vì ABCD là hình bình hành $\to CD=AB=\dfrac35DG$
$\to CG=DG-DC=DG-\dfrac35DG=\dfrac25DG$
$\to\dfrac{AB}{GC}=\dfrac{\dfrac35DG}{\dfrac25DG}=\dfrac32$
$\to\dfrac{AK}{KG}=\dfrac{AB}{CG}=\dfrac32,AB//CG$
$\to\dfrac{AK}{AK+KG}=\dfrac3{3+2}$
$\to\dfrac{AK}{AG}=\dfrac35\to AK=\dfrac35AG=\dfrac{32}{5}$
$\to EK=AK-AE=\dfrac{32}{5}-4=\dfrac{12}{5}$
b.Vì $AB//DG\to\dfrac{AB}{DG}=\dfrac{EB}{ED}$
Do $BK//AD\to\dfrac{BK}{AD}=\dfrac{BE}{ED}$
