Ta có: \(\left\{{}\begin{matrix}\widehat{ABC}=\widehat{ACB}\\\widehat{CBD}=\widehat{BCD}\end{matrix}\right.\) Cộng hai vế ta được: \(\widehat{ABD}=\widehat{ACD}\)
Xét \(\Delta ABD.và.\Delta ACD\) có:
\(\left\{{}\begin{matrix}AB=AC\\\widehat{ABD}=\widehat{ACD}\\BD=CD\end{matrix}\right.\Rightarrow\Delta ABD=\Delta ACD\left(c-g-c\right)\)
\(\Rightarrow\widehat{BOA}=\widehat{CDA}\)
Lại có: \(\widehat{BDA}+\widehat{CDA}=\widehat{BDA}=60^0\)
Mà: \(2\widehat{BDA}=60^0\)
\(\Rightarrow\widehat{BDA}=30^0\)
Vậy .............
