a) $\begin{cases}c⊥AD\\d⊥AD\end{cases}→c//d$
b) $c//d→\widehat{B_1}+\widehat{C_1}=180^\circ$ (trong cùng phía)
mà $\widehat{B_1}=60^\circ$
$→\widehat{C_1}=180^\circ-60^\circ=120^\circ$
c) $c//d→\begin{cases}\widehat{BCD}=\widehat{B_1}=60^\circ\\\widehat{A_1}=\widehat{C_1}=30^\circ\end{cases}→\widehat{ACB}=60^\circ-30^\circ=30^\circ$
Ta có: $\begin{cases}\widehat{BCA}=\widehat{A_1}\\\widehat{C_1}=\widehat{A_1}\end{cases}→\widehat{BCA}=\widehat{C_1}$
$→CA$ là tia phân giác $\widehat{BCD}$
