Trong \(\Delta ABE\) có \(BE=AE\)
\(\Rightarrow\Delta ABE\) cân tại \(E\).
\(\Rightarrow\widehat{ABE}=\widehat{BAE}\)
Trong \(\Delta ACE\) có\(CE=AE\)
\(\Rightarrow\Delta ACE\) cân tại \(E\)
\(\Rightarrow\widehat{ACE}=\widehat{CAE}\)
Ta có : \(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\)
Hay : \(\widehat{BAE}+\widehat{CAE}+\widehat{ABE}+\widehat{ACE}=180^o\)
\(\Leftrightarrow2.\widehat{BAE}+2.\widehat{CAE}=180^o\)
\(\Leftrightarrow2.\left(\widehat{BAE}+\widehat{CAE}\right)=180^o\)
\(\Leftrightarrow2.\widehat{BAC}=180^o\)
\(\Rightarrow\widehat{BAC}=90^o\)
Do đó : \(\Delta ABC\) có \(\widehat{BAC}=90^o\)
Nên \(\Delta ABC\) vuông tại \(A\)
