Question

Answer this .......... according to thererom of circleanswer only (a) part

Answer 1

(a) In the given figure,

`\angle ABO=x`

BD is the bisector of angle ABC, thus:

`\angle ABD=\angle DBC`

So, write as follows:

(i)

`\angle ABC=\angle ABD+\angle DBC`

`\begin{gathered} \angle ABC=\angle ABD+\angle ABD \\ \angle ABC=2\angle ABD \\ \angle ABC=2\angle ABO \\ \angle ABC=2x \end{gathered}`

Therefore,

(ii)

OB=OA=OD radii of the same circle.

As OB=OA then in triangle AOB,

`\angle ABO=\angle BAO=x`

BOD is a diameter of a circle. then by theorem of circle,

`\angle BAD=90^(\circ)`

thus,

`\angle OAD=\angle BAD-\angle OAB`

`\angle OAD=90^(\circ)-x`

OA=OD , in the tringle AOD,

`\angle OAD=\angle ODA=90^(\circ)-x`

In triangle AOD,

`\angle AOD+\angle OAD+\angle ODA=180^(\circ)`

`\begin{gathered} \angle AOD+90^(\circ)-x+90^(\circ)-x=180^(\circ) \\ \angle AOD+180^(\circ)-2x=180^(\circ) \\ \angle AOD=2x \end{gathered}`

Therefore,

`\begin{equation*} \angle AOD=2x \end{equation*}`

(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.

`\angle AOC=2\angle ABC`

`\begin{gathered} \angle AOC=2*2x \\ \angle AOC=4x \end{gathered}`

Therefore,

(iv)

OA=OD , thus tringle AOD is an isosceles triangle,

`\angle OAD=\angle ODA=90^(\circ)-x`

`\begin{gathered} \angle ADO=90^(\circ)-x \\ \angle ADB=90^(\circ)-x \end{gathered}`

Therefore,

`\begin{equation*} \angle ADB=90^(\circ)-x \end{equation*}`