Question

Answer 1

Xét ΔABC có AD/AB=AE/AC
nên ED//BC

Answer 2

Trên tia đối của ED lấy EF s/c ED=EF

Nối F với B

△ ADE=△ CFE(c.g.c)

\(\left\{{}\begin{matrix}EF=ED\\EC=AE\\\widehat{AED}=\widehat{FEC}\end{matrix}\right.\)

\(\Rightarrow\widehat{ADE}=\widehat{F},AD=FC=BD\left(...\right)\)

⇒ BD//FC

⇒ \(\widehat{DBC}=\widehat{BFC}\)

△ BDF=△FCB(c.g.c)

\(\left\{{}\begin{matrix}\widehat{DBC}=\widehat{BFC}\\BFchung\\BD=FC\end{matrix}\right.\)

\(\Rightarrow\widehat{B}=\widehat{F}=\widehat{ADE}\)

⇒ DE//BC