In the diagram, PAQ is the tangent to the circle at A. The line BCT is parallel to PAQ and ADT is a straight line.

1. Relate angle ADB to the angle between the tangent and chord AB.

$\angle PAB = \angle ADB$ (Alternate Segment Theorem)

2. Relate angle PAB to angle ABC using parallel lines.

$\angle PAB = \angle ABC$ (Alternate angles, $PAQ \parallel BCT$)

3. Relate angle CDT to angle ABC using cyclic quadrilateral properties.

$\angle CDT = \angle ABC$ (Exterior angle of cyclic quadrilateral $ABCD$)

4. Conclude the proof.

Since $\angle PAB = \angle ADB$ and $\angle PAB = \angle ABC$, we have $\angle ADB = \angle ABC$.
Since $\angle CDT = \angle ABC$, it follows that $\angle ADB = \angle CDT$. (Shown)

1. State the first pair of equal angles from part (a).

From part (a), $\angle ADB = \angle CDT$.

2. Find a second pair of equal angles using alternate segment theorem and parallel lines.

Let $\angle QAD = x$.
By the alternate segment theorem, $\angle ABD = \angle QAD = x$.
Since $PAQ \parallel BCT$, alternate angles are equal, so $\angle CTD = \angle QAD = x$.
Therefore, $\angle ABD = \angle CTD$.

3. Conclude similarity using AA criterion.

In $\triangle TCD$ and $\triangle BAD$:

1. $\angle CDT = \angle ADB$ (proven in (a))
2. $\angle CTD = \angle ABD$ (proven above)
   Since two pairs of corresponding angles are equal, $\triangle TCD$ and $\triangle BAD$ are similar (AA similarity).