Answer :
Solution:
(a) In the given figure,
[tex]\angle ABO=x[/tex]BD is the bisector of angle ABC, thus:
[tex]\angle ABD=\angle DBC[/tex]So, write as follows:
(i)
[tex]\angle ABC=\angle ABD+\angle DBC[/tex][tex]\begin{gathered} \angle ABC=\angle ABD+\angle ABD \\ \angle ABC=2\angle ABD \\ \angle ABC=2\angle ABO \\ \angle ABC=2x \end{gathered}[/tex]Therefore,
[tex]\operatorname{\angle}ABC=2x[/tex](ii)
OB=OA=OD radii of the same circle.
As OB=OA then in triangle AOB,
[tex]\angle ABO=\angle BAO=x[/tex]BOD is a diameter of a circle. then by theorem of circle,
[tex]\angle BAD=90^{\circ}[/tex]thus,
[tex]\angle OAD=\angle BAD-\angle OAB[/tex][tex]\angle OAD=90^{\circ}-x[/tex]OA=OD , in the tringle AOD,
[tex]\angle OAD=\angle ODA=90^{\circ}-x[/tex]In triangle AOD,
[tex]\angle AOD+\angle OAD+\angle ODA=180^{\circ}[/tex][tex]\begin{gathered} \angle AOD+90^{\circ}-x+90^{\circ}-x=180^{\circ} \\ \angle AOD+180^{\circ}-2x=180^{\circ} \\ \angle AOD=2x \end{gathered}[/tex]Therefore,
[tex]\begin{equation*} \angle AOD=2x \end{equation*}[/tex](iii) By the theorem of circle, the angle formed at the center of the circle is twice the angle formed at the circumference of the circle with the same base.
[tex]\angle AOC=2\angle ABC[/tex][tex]\begin{gathered} \angle AOC=2\times2x \\ \angle AOC=4x \end{gathered}[/tex]Therefore,
[tex]\operatorname{\angle}AOC=4x[/tex](iv)
OA=OD , thus tringle AOD is an isosceles triangle,
[tex]\angle OAD=\angle ODA=90^{\circ}-x[/tex][tex]\begin{gathered} \angle ADO=90^{\circ}-x \\ \angle ADB=90^{\circ}-x \end{gathered}[/tex]Therefore,
[tex]\begin{equation*} \angle ADB=90^{\circ}-x \end{equation*}[/tex]