Giải:
Xét \(\Delta BIC\) có: \(\widehat{BIC}+\widehat{B_1}+\widehat{C_1}=180^o\)
\(\Rightarrow135^o+\widehat{B_1}+\widehat{C_1}=180^o\)
\(\Rightarrow\widehat{B_1}+\widehat{C_1}=45^o\)
\(\Rightarrow\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}=45^o\)
\(\Rightarrow\frac{1}{2}\left(\widehat{B}+\widehat{C}\right)=45^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=90^o\)
Trong \(\Delta ABC\) có: \(\widehat{B}+\widehat{C}=90^o\Rightarrow\widehat{A}=90^o\)
\(\Rightarrow\Delta ABC\) vuông tại \(\widehat{A}\)
Vậy...