a) ABCD là hình thang cân $\Rightarrow \widehat{D}=\widehat{C};\widehat{A}=\widehat{B}$

• Xét ∆ACD có: $\widehat{ADC}+\widehat{ACD}=180°-\widehat{DAC}=180°-90°=90°$

• Xét ∆ABC có: AB = BC => ∆ABC cân tại B

$\Rightarrow \widehat{BAC}=\widehat{BCA}$

Ta có: $\widehat{DAB}=\widehat{DAC}+\widehat{CAB}$

$$\begin{gathered} =90°+\widehat{CAB}=\widehat{ADC}+\widehat{ACD}+\widehat{ACB} \\ =\widehat{BCD}+\widehat{BCD}=2\widehat{BCD} \end{gathered}$$

Mà $\widehat{DAB}+\widehat{BCD}=180°\Rightarrow 2\widehat{BCD}+\widehat{BCD}=180°$

$$\begin{gathered} \Rightarrow 3\widehat{BCD}=180°\Rightarrow \widehat{BCD}=60° \\ \Rightarrow \widehat{C}=\widehat{D}=60° \\ \Rightarrow \widehat{A}=\widehat{B}=180°-\widehat{C}=180°-60°=120° \end{gathered}$$