Need help with inscribed angles If M

Given:

In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and [tex]m\angle BCD=30^\circ[/tex].

To find:

The [tex]m\angle AOB[/tex].

Solution:

If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

We have, DC is parallel to BA and BC is the transversal line.

[tex]\angle OBA\cong \angle BCD[/tex]        [Alternate interior angles]

[tex]m\angle OBA=m\angle BCD[/tex]

[tex]m\angle OBA=30^\circ[/tex]

In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.

We know that base angles of an isosceles triangle are congruent.

[tex]\angle OAB\cong \angle OBA[/tex]      [Base angles of an isosceles triangle]

[tex]m\angle OAB=m\angle OBA[/tex]

[tex]m\angle OAB=30^\circ[/tex]

In triangle AOB,

[tex]m\angle OAB+m\angle OBA+m\angle AOB=180^\circ[/tex]

[tex]30^\circ+30^\circ+m\angle AOB=180^\circ[/tex]

[tex]60^\circ+m\angle AOB=180^\circ[/tex]

[tex]m\angle AOB=180^\circ-60^\circ[/tex]

[tex]m\angle AOB=120^\circ[/tex]

Therefore, the measure of angle AOB is 120 degrees.

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