In the following figure, O is the center of the circle such that ∠ABC

SOLUTION

Join AC and AO, then ∠AOC = 2 x 30° = 60° (Angle made by a chord at centre of circle is double of the angle made by the same chord on circumference)

Now, AO = CO ⇒ ∠OAC = ∠OCA = 60° 

∠ACB = ∠ACO +∠OCB = 60° + 20° = 80°
In Triangle ABC;
⇒ ∠CAB = 180° – (∠ACB +∠ABC) =180° – (80° + 30°) = 70°
Also, ∠CDA is external angle of triangle BDC;
⇒ ∠CDA = ∠DCB + ∠B = 40° + 30° = 70°
Now, From triangle OAC;
OAC is an equilateral Triangle; so 
⇒ CA = CO
also, in triangle CDA;
∠CDA = ∠CAD = 70°
so, CD = CO
this makes-
⇒ ∠CDO = ∠COD (in Triangle COD)