1) Xét ΔBCE và ΔADC có:
\(\frac{BC}{AD}=\frac{BE}{AC}=\frac{1}{3};EBC=\widehat{CAD}=90^0\)
⇒ΔBCE~ΔADC (c.g.c)
2) Từ phần 1) suy ra \(\widehat{ECB}=\widehat{CDA}\Rightarrow\widehat{ECB}+\widehat{DCA}=\widehat{CDA}+\widehat{DCA}=90^0\)
⇒\(\widehat{DCE}=90^0\)