Xét \(\Delta ABC\) có \(\widehat{DAC}\) là goác ngoài của \(\Delta\) tại \(A\)
\(\Rightarrow\widehat{DAC}=\widehat{ABC}+\widehat{ACB}\left(t/c\right)\)
Vì \(\Delta ABC\) cân tại \(A\)
\(\Rightarrow\left\{{}\begin{matrix}AB=AC\\\widehat{ABC}=\widehat{ACB}\left(t/c\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{DAC}=2.\widehat{ACB}\)
Vì \(\left\{{}\begin{matrix}AB=AC\left(cmt\right)\\AB=AD\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow AC=AD\)
\(\Rightarrow\Delta DAC\) cân tại \(A\)
\(\Rightarrow\widehat{ADC}=\widehat{ACD}\left(t/c\right)\)
Xét \(\Delta DAC\) có \(\widehat{BAC}\) là goác ngoài của \(\Delta\) tại \(A\)
\(\Rightarrow\widehat{BAC}=\widehat{ADC}+\widehat{ACD}\left(t/c\right)\)
Mà \(\widehat{ADC}=\widehat{ACD}\left(cmt\right)\)
\(\Rightarrow\widehat{BAC}=2.\widehat{ACD}\)
Vì \(\widehat{DAC}+\widehat{BAC}=180^o\) (2 góc kề bù)
\(\Leftrightarrow2.\widehat{ACB}+2.\widehat{ACD}=180^o\)
\(\Leftrightarrow2.\left(\widehat{ACB}+\widehat{ACD}\right)=180^o\)
\(\Leftrightarrow2.\widehat{BCD}=180^o\)
\(\Leftrightarrow\widehat{BCD}=90^o\)
\(\Rightarrow\Delta BCD\) là \(\Delta\) vuông.
