a/ \(DE//BC\\→\dfrac{AE}{AC}=\dfrac{DE}{BC}(ĐL\,\, Thales)\\→\dfrac{5}{15}=\dfrac{DE}{18}\\→DE=6\)
b/ \(DE//BC\\→\dfrac{AE}{EC}=\dfrac{AD}{DB}(ĐL\,\, Thales)\\→\dfrac{AE}{AC-AE}=\dfrac{AD}{DB}\\→\dfrac{5}{10}=\dfrac{AD}{DB}\\→\dfrac{1}{2}=\dfrac{AD}{DB}\)