$\text{Vì $\Delta$ABC cân tại A (gt) $\rightarrow$ $\widehat{ACB}$ = $\widehat{ABC}$}$

$\text{Vì $\Delta$BCD là tam giác đều (gt) $\rightarrow$ $\widehat{DBC}$ = $\widehat{BDC}$ = $\widehat{DCB}$ = $60^o$}$

$\text{Ta có: }$

$\text{$\widehat{ABD}$ = $\widehat{ABC}$ + $\widehat{CBD}$}$ 

$\text{$\widehat{ACD}$ = $\widehat{ACB}$ + $\widehat{DCB}$}$ 

$\text{$\Rightarrow$ $\widehat{ABD}$ = $\widehat{ACD}$}$

$\text{Xét $\Delta$ABD và $\Delta$ACD có:}$

$\text{AD: cạnh chung}$

$\text{ $\widehat{ABD}$ = $\widehat{ACD}$}$

$\text{AB = AC}$

$\Rightarrow$ $\text{$\Delta$ABD = $\Delta$ACD (c.g.c)}$

$\text{$\Rightarrow$ $\widehat{BDA}$ = $\widehat{CDA}$ = $\dfrac{\widehat{BDC}}{2}$ = $\dfrac{60^o}{2}$ = $30^o$}$

$\text{Vậy $\widehat{BDA}$ = $30^o$ }$