[tex]\it\dfrac{AB}{DB} =\dfrac{3}{2} \Rightarrow AB=3k,\ \ DB=2k,\ \ AD=AB-DB=3k-2k=k,\ (k\in\mathbb{Q_+}) \\ \\ \\ DE||BC\ \stackrel{T.f.a.}{\Longrightarrow}\ \ \Delta ADE\sim \Delta ABC \Rightarrow \dfrac{AD}{AB}=\dfrac{DE}{BC}=\dfrac{AE}{AC}\\ \\ \\ \dfrac{AD}{AB}=\dfrac{DE}{BC} \Rightarrow \dfrac{k}{3k}=\dfrac{DE}{20} \Rightarrow \dfrac{1}{3}=\dfrac{DE}{20} \Rightarrow DE=\dfrac{20}{3}=6\dfrac{2}{3}\ cm[/tex]
[tex]\it \dfrac{AD}{AB}=\dfrac{AE}{AC} \Rightarrow \dfrac{k}{3k}=\dfrac{AE}{15} \Rightarrow AE=\dfrac{15k}{3k} \Rightarrow AE = 5\ cm\\ \\ EC=AC-AE=15-5=10\ cm[/tex]
Raportul de asemănare a triunghiurilor ADE și ABC este k/3k = 1/3.
Raportul ariilor celor două triunghiuri este egal cu pătratul raportului
de asemănare, deci:
[tex]\it \dfrac{\mathcal{A}_{ADE}}{\mathcal{A}_{ABC}}=\Big(\dfrac{1}{3}\Big)^2=\dfrac{1}{9}[/tex]
