\(S_{ADFE}=S_{ABC}-\left(S_{BDF}+S_{CFE}\right)=\dfrac{AC.AB}{2}-\left(\dfrac{EC.FE}{2}+\dfrac{BD.DF}{2}\right)\)

Do FE=AD và DF=AE nên

\(S_{ADFE}=\dfrac{\left(AE+EC\right)\left(AD+BD\right)}{2}-\left(\dfrac{EC.AD}{2}+\dfrac{BD.AE}{2}\right)\)

\(\Rightarrow2.S_{ADFE}=AE.AD+BD.AE+EC.AD+EC.BD-EC.AD-BD.AE=AE.AD+EC.BD=S_{ADFE}+EC.BD\)

\(\Rightarrow S_{ADFE}=EC.BD=25.20=500cm^2\)