Giải thích các bước giải:
Ta có : $AD, CE$ là phân giác của tam giác
$\to\dfrac{EA}{EB}=\dfrac{CA}{CB}\to\dfrac{AE}{AE+EB}=\dfrac{CA}{CB+CA}\to\dfrac{AE}{AB}=\dfrac{AC}{BC+AC}$
$\dfrac{DB}{DC}=\dfrac{AB}{AC}\to\dfrac{BD+DC}{DC}=\dfrac{AB+AC}{AC}\to\dfrac{BC}{DC}=\dfrac{AB+AC}{AC}$
$\to \dfrac{AE}{AB}.\dfrac{BC}{DC}=\dfrac{AC}{BC+AC}.\dfrac{AB+AC}{AC}$
$\to \dfrac{AE}{AB}{DC}.\dfrac{BC}{AB}=\dfrac{AB+AC}{BC+AC}$
$\to \dfrac{AE}{DC}=\dfrac{AB+AC}{BC+AC}.\dfrac{AB}{BC}$
VÌ $AB>BC\to \dfrac{AB+AC}{BC+AC}.\dfrac{AB}{BC}>1\to AE>DC$
