Ta có: Tam giác ACD đều => $\widehat{ACD}=\widehat{ADC}=\widehat{CAD}=60^{\circ}$

Tam giác ABC cân tại C

=> $\widehat{ABC}$ = $\widehat{BAC}$ =$\frac{\widehat{ABC} +\widehat{BAC}}{2}$ = $\frac{180^{\circ}−\widehat{ACB}}{2}=\frac{\widehat{ACD}}{2}=30^{\circ}$

Tam giác ADE cân tại đỉnh D nên:

$\widehat{AED}=\widehat{DAE}=\frac{AED+DAE}{2}=\frac{180^{\circ}−\widehat{ADE}}{2}=\frac{\widehat{ADC}}{2}=30^{\circ}$

=>$\widehat{ABE}=\widehat{ABC}=30^{\circ}$

$\widehat{AEB}=\widehat{AED}=30^{\circ}$

$\widehat{BAE}=180^{\circ}−\widehat{ABE}−\widehat{AEB}=120^{\circ}$