Xét ΔABC có \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\left(=2\right)\)
nên DE//BC
Xét ΔABC có DE//BC
nên \(\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac{2}{3}\)
Xét ΔIDE và ΔICB có
\(\widehat{IDE}=\widehat{ICB}\)(hai góc so le trong, DE//CB)
\(\widehat{DIE}=\widehat{CIB}\)(hai góc đối đỉnh)
Do đó: ΔIDE đồng dạng với ΔICB
=>\(\dfrac{ID}{IC}=\dfrac{IE}{IB}=\dfrac{DE}{BC}=\dfrac{2}{3}\)
Vì AE=2/3AC
nên \(S_{AEB}=\dfrac{2}{3}\cdot S_{ABC}\)
IE/IB=2/3
=>\(\dfrac{IB}{IE}=\dfrac{3}{2}\)
=>\(\dfrac{IB+IE}{IE}=\dfrac{3+2}{2}\)
=>\(\dfrac{BE}{IE}=\dfrac{5}{2}\)
=>\(\dfrac{IE}{BE}=\dfrac{2}{5}\)
=>\(S_{AIE}=\dfrac{2}{5}\cdot S_{ABE}=\dfrac{2}{5}\cdot\dfrac{2}{3}\cdot S_{ABC}=\dfrac{4}{15}\cdot S_{ABC}\)(1)
Vì BD=1/3AB
nên \(S_{BDC}=\dfrac{1}{3}\cdot S_{ABC}\)
\(\dfrac{ID}{IC}=\dfrac{2}{3}\)
=>\(\dfrac{IC}{ID}=\dfrac{3}{2}\)
=>\(\dfrac{IC+ID}{ID}=\dfrac{3+2}{2}\)
=>\(\dfrac{CD}{ID}=\dfrac{5}{2}\)
=>\(\dfrac{DI}{DC}=\dfrac{2}{5}\)
=>\(S_{DIB}=\dfrac{2}{5}\cdot S_{DBC}=\dfrac{2}{5}\cdot\dfrac{1}{3}\cdot S_{ABC}=\dfrac{2}{15}\cdot S_{ABC}\)
=>\(S_{IAE}=2\cdot S_{DIB}\)
