a, Vì \(\left\{{}\begin{matrix}m\perp AB\\n\perp AB\end{matrix}\right.\) nên m//n
b, Vì m//n nên \(\left\{{}\begin{matrix}\widehat{D_1}=\widehat{C_4}=110^0\left(so.le.trong\right)\\\widehat{D_1}=\widehat{C_2}=110^0\left(đồng.vị\right)\end{matrix}\right.\)
Ta có \(\widehat{C_1}+\widehat{C_2}=180^0\left(kề.bù\right)\Rightarrow\widehat{C_1}=180^0-110^0=70^0\)
Ta có \(\widehat{C_1}=\widehat{C_3}=70^0\left(đối.đỉnh\right)\)
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