The problem you're working on involves geometry of a circle, and we need to find two specific angles: $\angle BDC$ and $\angle BTD$.

Let's break down the given information:

1. Circle Information:

   • $A, B, C, D$ are points on the circle.
   • $AB$ is the diameter of the circle, which means $\angle ACB = 90^\circ$ (since an angle subtended by a diameter in a circle is always a right angle).
   • $DC$ is parallel to $AB$.

2. Given Angles:

   • $\angle BAD = 70^\circ$.
   • $\angle ACB = 20^\circ$.

Part (a) - Calculate the size of angle $\angle BDC$.

Since $DC \parallel AB$, and given that $AB$ is the diameter of the circle, we can use properties of parallel lines and cyclic quadrilaterals.

In the cyclic quadrilateral $ABCD$, the opposite angles should sum up to 180°. Thus, $\angle BDC$ will be supplementary to $\angle BAD$, because both are angles on the same side of the line $BC$. Therefore:

$\angle BDC = 180^\circ - \angle BAD = 180^\circ - 70^\circ = 110^\circ.$

Part (b) - Calculate the size of angle $\angle BTD$.

This is a tangent-secant angle. The angle formed between a tangent and a chord through the point of contact is equal to the angle subtended by the chord at the opposite side of the circle. In this case:

$\angle BTD = \angle BDC = 110^\circ.$

Final Answers:

• $\angle BDC = 110^\circ$.
• $\angle BTD = 110^\circ$.

Let me know if you'd like more details or further explanations!

Here are some related questions you might find helpful:

1. How does the angle subtended by a diameter of a circle relate to the angles in the circle?
2. What is the property of angles in a cyclic quadrilateral?
3. Can you explain how parallel lines impact the angles in a circle?
4. What is the definition of a tangent and how does it affect the angles?
5. How can you calculate angles using properties of cyclic quadrilaterals?

Tip: When working with circle geometry, always remember the key properties, like the angle in a semicircle being a right angle, and the fact that opposite angles in a cyclic quadrilateral sum to 180°.