Properties of circle part-2

Join AC
then $\angle CAD = 90 ^ \circ$ (because CD is diameter)
Then $\angle CAO = 70 ^ \circ$ .
$\bigtriangleup$ CAO is an isosceles $\bigtriangleup$ because OA and OC is radius.
$\angle OCA$ = $\angle OAC$
$\angle OCA = 70 ^ \circ$ Then $\angle COA = 40 ^ \circ$ (Angle sum property)
$\angle COA$ = $\angle OAB$ (Alternate interior angle)

$\angle OAB$ = $\angle OBA$ ( OA = OB )
So, $\angle ABC = 100 ^ \circ$

As $\bigtriangleup OAD$ is isosceles, we have $\angle ODA = 20^{\circ}$ and, as $\overline{CD} \parallel \overline{AB}$ , we have $\angle DAB = 20^{\circ}$ .

Therefore, as $\bigtriangleup OAB$ is isosceles, we have $\angle OBA = 40^{\circ}$ and, hence, $\angle AOB = 100^{\circ}$

Just focus on triangle DOA, since it is an isosceles triangle, angle A and D are equal to 20 deg, and angle O is 140. OA and OB are also equal, so it means that it also have equal angles. So make a perpendicular line from point O to line AB, from that you made a 90 deg angle, subtract 90 from 140 then multiply it by two, then youll get 100 degrees

Given that,

CD||AB

<DAO=20 degrees

In this circle,

radius,OA=OB=OC=OD

In triangle AOD

OA=OD

so,<ODA=<OAD

therefore,<ODA=20 degrees (given that,angle DAO=20 degrees)

again, as CD||AB

so, <CDA=<DAB (as,they are alternative angles)

=> <ODA=<DAB

so, <DAB=20 degrees

now, <OAD=20 degrees,<DAB=20 degrees

so,<OAB=<OAD+<DAB =(20+20) degrees =40 degrees

again,in triangle OAB

OA=OB (as they are radius)

so,<OBA=<OAB

therefore,<OBA=40 degrees

in triangle OAB

<OAB+<OBA+<AOB=180 degrees

=> <AOB+80 degrees=180 degrees

so,<AOB=100 degrees