I have a hard circle theorem problem:

a.

BAD  = 70°

Since BDA  has its endpoints on a diameter, it measures 90°

And ABD  =   180 - 70  - 90  =  20°

Since BD is a trasnsversal cutting two parallel lines....BDC  =  ABD.....therefore, BDC  = 20°

b.  

Note..... BCD  is supplemental to BAD    =  110°

And angle DBC   =  180  -  BDC - DCB =  180  - 20  - 110  =  50°

And minor arc DC  =  2DBC  =  100°

And angle CDT  =  (1/2) minor arc DC  = 50°

And by the experior angle theorem, angle BCD  =  angle CDT  + angle CTD  ...so..

BCD  = CDT  + CTD

110  =  50   +  angle CTD    .....subtract 50 from both sides

60°  = angle CTD  = angle BTD