Ta có \(\Delta_vABD\sim\Delta_vACE\) (chung góc A)
\(\Rightarrow\dfrac{AB}{AC}=\dfrac{AD}{AE}\) \(\Rightarrow\dfrac{AB}{AD}=\dfrac{AC}{AE}\)
Xét hai tam giác ADE và ABC có: \(\left\{{}\begin{matrix}\widehat{A}-chung\\\dfrac{AB}{AD}=\dfrac{AC}{AE}\end{matrix}\right.\)
\(\Rightarrow\Delta ADE\sim\Delta ABC\) (c.g.c)
\(\Rightarrow\dfrac{ED}{BC}=\dfrac{AD}{AB}=cosA=cos45^0=\dfrac{\sqrt{2}}{2}\)