Trong tam giác ABC lấy I sao cho BIC đều.
Khi đó [tex]BC=IB=IC;\widehat{IBC}=\widehat{ICB}=60^o[/tex]
Ta có:[tex]\widehat{BAC}=20^o\Rightarrow \widehat{ABC}=\widehat{ACB}=80^o\Rightarrow \widehat{IBC}=20^o[/tex]
Xét 2 tam giác ABI và ACD.
[tex]\left.\begin{matrix} IB=AD(=BC)\\ \widehat{IBC}=\widehat{CAD}(=20^o)\\ AB=AC \end{matrix}\right\}\Rightarrow \Delta IBC=\Delta DAC(c-g-c)\Rightarrow \widehat{CDA}=\widehat{CIB}=\frac{360^o-\widehat{BIC}}{2}=\frac{360^o-60^o}{2}=150^o[/tex]