The points A, B, C and D lie on the circumference of a circle centre O. AB and DC are parallel. BD is a straight line passing through O and BDC = 38°. TV is a tangent of the circle at D. Find the size of (a) TDS, (b) CBD, (c) ABD.

Since BD is a straight line passing through the center O, BD is the diameter of the circle.

The angle subtended by a diameter at any point on the circumference is $90^\circ$. Therefore, $\angle BCD = 90^\circ$.

In $\triangle BCD$, the sum of angles is $180^\circ$:

Given $\angle BDC = 38^\circ$ and $\angle BCD = 90^\circ$:

The angle between the tangent TV and the chord DC (i.e., $\angle TDC$) is equal to the angle in the alternate segment subtended by the chord DC (i.e., $\angle DBC$).
Therefore, $\angle TDC = \angle DBC = 52^\circ$.
The size of $\angle TDC$ is $\boxed{\text{52}^\circ}$.

We are given that AB is parallel to DC ($AB \parallel DC$).

BD is a transversal line intersecting the parallel lines AB and DC. Therefore, $\angle ABD$ and $\angle BDC$ are alternate interior angles.

Alternate interior angles are equal when lines are parallel.

Given $\angle BDC = 38^\circ$:

The size of $\angle ABD$ is $\boxed{\text{38}^\circ}$.