Circles — Maths STD 9 — Question

$ABCD$ is a cyclic Quadrilateral.
Consider $\triangle\text{ABD}$ and $\triangle\text{ABC}.$
Both are on the same base $AB$ and $\angle\text{ADB}$ and $\angle\text{ACB}$ are the angles in the same segment $AB.$
$\Rightarrow\angle\text{ADB}=\angle\text{ACB}=30^\circ$
$\Rightarrow\angle\text{BCD}=80^\circ+30^\circ=110^\circ$
In a cyclic Quadrilateral, sum of opposite angles is $180^\circ $
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$