Giải thích các bước giải:
a.Vì $BD$ là phân giác góc $B$
$\to \dfrac{DA}{DC}=\dfrac{BA}{BC}=\dfrac32$
$\to \dfrac{DA}{DA+DC}=\dfrac3{3+2}$
$\to \dfrac{AD}{AC}=\dfrac35$
$\to AD=\dfrac35AC=\dfrac{18}{5}$
Tương tự ta có $\dfrac{AE}{AB}=\dfrac35\to \dfrac{AE}{AB}=\dfrac{AD}{AC}\to DE//BC$
$\to\dfrac{DE}{BC}=\dfrac{AD}{AC}=\dfrac35$
$\to DE=\dfrac35BC=\dfrac{12}5$
b.Từ câu a
$\to\dfrac{AE}{AB}=\dfrac{AD}{AC}$
$\to\dfrac{AE}{AC}=\dfrac{AD}{AB}$ vì $AB=AC$
$\to\Delta ABD\sim\Delta AEC(c.g.c)$
c.Do $DE//BC$ (câu a)
$\to \dfrac{IE}{IC}=\dfrac{ID}{IB}$
Mà $\widehat{EIB}=\widehat{DIC}$
$\to \Delta EIB\sim\Delta DIC(c.g.c)$
$\to\dfrac{EI}{DI}=\dfrac{EB}{CD}$
$\to IE.CD=ID.BE$
d.Do $DE//BC\to \widehat{ADE}=\widehat{ACB}$
$\to \Delta ADE\sim\Delta ACB(g.g)$
$\to \dfrac{S_{ADE}}{S_{ABC}}=(\dfrac{AD}{AC})^2=\dfrac9{25}$
$\to S_{ADE}=\dfrac9{25}S_{ABC}=\dfrac{108}5$
