Ta có: $\text{ΔABCcân tại A}$

$\text{nên: $\widehat{ABC}$ = $\widehat{ACB}$ = $50^o$}$

⇒ $\text{$\widehat{BAC}$ = $80^o$}$

$\text{Ta có: AB= AC (ΔABC vân tại A)}$

$\text{mà CD= AB (gt)}$

⇒ $\text{CD= AC}$

⇒ $\text{ΔCAD cân tại C}$

⇒ $\text{$\widehat{CAD}$ = $\widehat{CDA}$ = $\dfrac{180^o-(180^o-50^o)}{2}$}$

⇒ $\text{ $\widehat{CAD}$ = $25^o$}$

Ta có: $\text{$\widehat{BAD}$ = $\widehat{BAC}$ + $\widehat{CAD}$}$

                    $\text{= $80^o$ + $25^o$}$

                    $\text{= $105^o$}$

    $\text{Vậy...}$

$\text{#Kudo}$

$\text{Có: ΔABC cân tại A (gt) nên:}$

$\text{⇒ $\widehat{ABC}$ = $\widehat{ACB}$ = $\dfrac{180^o - \widehat{BAC}}{2}$ = $50^o$}$

$\text{⇒ $\widehat{BAC}$ = $180^o$ - 2 . $50^o$ = $180^o$ - $100^o$ - $80^o$}$

$\text{Có: $\widehat{ACD}$ + $\widehat{ACB}$ = $180^o$ (Tính chất 2 góc kề bù)}$

$\text{$\widehat{ACD}$ + $50^o$ = $180^o$ (Thay số)}$

$\text{$\widehat{ACD}$ = $130^o$}$

$\text{Có: CD = AB (gt); AB = AC (ΔABC cân tại A)}$

$\text{⇒ AC = CD}$

$\text{Xét ΔACD, có: AC = CD (cmt) nên:}$

$\text{⇒ ΔACD cân tại C (dhnb)}$

$\text{⇒ $\widehat{CAD}$ = $\widehat{CDA}$ = $\dfrac{180^o - \widehat{ACD}}{2}$ = $\dfrac{180^o - 130^o}{2}$ = $25^o$}$

$\text{Có: $\widehat{BAD}$ = $\widehat{BAC}$ + $\widehat{CAD}$ (Tính chất cộng góc)}$

$\text{$\widehat{BAD}$ = $80^o$ + $25^o$ (Thay số)}$

$\text{$\widehat{BAD}$ = $105^o$ }$

$\textit{Ha1zzz}$