Tam giác ACD đều, nên $\widehat{ACD}=\widehat{ADC}=\widehat{CAD}=60^{\circ}$.
Tam giác ABC cân tại đỉnh C nên:
$\widehat{ABC}=\widehat{BAC}=\frac{\widehat{ABC}+\widehat{BAC}}{2}=\frac{180^{\circ}-\widehat{ACB}}{2}=\frac{\widehat{ACD}}{2}=30^{\circ}$
Tam giác ADE cân tại đỉnh D nên:
$\widehat{AED}=\widehat{DAE}=\frac{\widehat{AED}+\widehat{DAE}}{2}=\frac{180^{\circ}-\widehat{ADE}}{2}=\frac{\widehat{ADC}}{2}=30^{\circ}$
Do vậy, ta có:
$\widehat{ABE}=\widehat{ABC}=30^{\circ},\widehat{AEB}=\widehat{AED}=30^{\circ},\widehat{BAE}=180^{\circ}-\widehat{ABE}-\widehat{AEB}=120^{\circ}$