Question

In the adjoining figure, $ABCD$ is a cyclic quadrilateral in which $\angle BCD={100}^o$ and $\angle ABD={50}^o$. Find $\angle ADB$.

Answer 1

It is given that $ABCD$ is a cyclic quadrilateral

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

$\angle A+\angle C={180}^o$

On further calculation

$\angle A={180}^o-{100}^o$

By subtraction

$\angle A={80}^o$

Consider $△ABD$

Using the angle sum property

$\angle A+\angle ABD+\angle ADB={180}^o$

By substituting the values

${80}^o+{50}^o+\angle ADB={180}^o$

On further calculation

$\angle ADB={180}^o-{80}^o-{50}^o$

By subtraction

$\angle ADB={180}^o-{130}^o$

So we get

$\angle ADB={50}^o$

Therefore, $\angle ADB={50}^o$.