B, ta co: \(\widehat{ACB}+\widehat{ACE}=180^0\left(kb\right)\)
\(\widehat{ABC}+\widehat{ABD}=180^0\left(kb\right)\)
ma \(\widehat{ABC}=\widehat{ACB}\) (ABC can tai A)
\(\Rightarrow\widehat{ABD}=\widehat{ACE}\)
A, Xet \(\Delta ABD\) va \(\Delta ACE\)
\(\widehat{ABD}=\widehat{ACE}\left(cmt\right)\)
AB= AC (ABC can tai A)
BD= CE (gt)
\(\Rightarrow\Delta ABD=\Delta ACE\left(cgc\right)\)
\(\Rightarrow AD=AE\) (2 canh tuong ung)
\(\Rightarrow\Delta ADE\) can tai A