$\text{Có: ΔABC là tam giác đều (gt)}$
⇒ $\text{$\widehat{BAC}$ = $60^{o}$ (tc Δ đều)}$
$\text{Xét ΔADC có:}$
$\text{$\widehat{DAC}$ + $\widehat{DCA}$ + $\widehat{ADC}$ = $180^{o}$ (đl tổng 3 góc Δ)}$
⇒ $\text{$\widehat{DAC}$ + $90^{o}$ + $\widehat{ADC}$ = $180^{o}$}$
⇒ $\text{$\widehat{DAC}$ + $\widehat{ADC}$ = $90^{o}$}$
$\text{mà $\widehat{DAC}$ = $\widehat{ADC}$ (ΔACD cân tại C)}$
⇒ $\text{$\widehat{DAC}$ = $\widehat{ADC}$ = $45^{o}$}$
$\text{Có: $\widehat{BAD}$ = $\widehat{DAC}$ + $\widehat{BAC}$}$
⇒ $\text{$\widehat{BAD}$ = $45^{o}$ + $60^{o}$}$
⇒ $\text{$\widehat{BAD}$ = $105^{o}$}$
