Question
In the adjoining figure, ABCD is a cyclic quadrilateral in which $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=50^\circ.$ Find $\angle\text{ADB}.$
✓
Answer
ABCD is a cyclic quadrilateral.$\therefore\ \angle\text{A}+\angle\text{C}=180^\circ$ [Opposite of a cyclic quadrilateral are supplementary]
$\Rightarrow\ \angle\text{A}+100^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}=180^\circ-100^\circ=80^\circ$
Now, in $\triangle\text{ABD},$ we have:$\angle\text{A}+\angle\text{ABD}+\angle\text{ADB}=180^\circ$
$\Rightarrow\ 80^\circ+50^\circ+\angle\text{ADB}=180^\circ$
$\Rightarrow\ \angle\text{ADB}=180^\circ-130^\circ=50^\circ$
$\therefore\ \angle\text{ADB}=50^\circ$
$\Rightarrow\ \angle\text{A}+100^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}=180^\circ-100^\circ=80^\circ$
Now, in $\triangle\text{ABD},$ we have:$\angle\text{A}+\angle\text{ABD}+\angle\text{ADB}=180^\circ$$\Rightarrow\ 80^\circ+50^\circ+\angle\text{ADB}=180^\circ$
$\Rightarrow\ \angle\text{ADB}=180^\circ-130^\circ=50^\circ$
$\therefore\ \angle\text{ADB}=50^\circ$
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