Question

In the given figure, $AD$ is a diameter, $O$ is the centre of the circle, $AD$ is parallel to $BC$ and $\angle CBD={32}^o$
Find:
(i) $\angle OBD$
(ii) $\angle AOB$
(iii) $\angle BED$

Answer 1

As $AD\parallel BC$ and $DB$ is a transversal.

$\Rightarrow$ $\angle ODB=\angle DBC$ [ Alternate angles ]

$\therefore$ $\angle ODB={32}^o$

In $△OBD$

$OB=OD$ [ Radius of a circle ]

$\therefore$ $△OBD$ is isosceles triangle.

$\Rightarrow$ $\angle OBD=\angle ODB$ [ Base angles are equal in isosceles triangle ]

$\therefore$ $\angle OBD={32}^o$

$\angle OBD+\angle ODB+\angle BOD={180}^o$ [ Sum of angles of triangle is ${180}^o.$ ]

$\Rightarrow$ ${32}^o+{32}^o+\angle BOD={180}^o$

$\Rightarrow$ ${64}^o+\angle BOD={180}^o$

$\Rightarrow$ $\angle BOD={116}^o$

$\Rightarrow$ $\angle BOD+\angle AOB={180}^o$ [ Linear pair ]

$\Rightarrow$ ${116}^o+\angle AOB={180}^o$

$\therefore$ $\angle AOB={64}^o$

In $△AOB,AO=OB$, hence its an isosceles triangle

$\Rightarrow$ $\angle OAB=\angle OBA$

Now, $\angle AOB+\angle OAB+\angle OBA={180}^o$

$\Rightarrow$ ${64}^o+\angle OAB+\angle OAB={180}^o$

$\Rightarrow$ $2\angle OAB={116}^o$

$\therefore$ $\angle OAB={58}^o$

$\Rightarrow$ $\angle OAB=\angle BED$ [ Angles subtended by the same chord $BD$ ]

$\therefore$ $\angle BED={58}^o.$

$\left(i\right)$ $\angle OBD={32}^o$

$\left(ii\right)$ $\angle AOB={64}^o$

$\left(iii\right)$ $\angle BED={58}^o.$