The diagram shows a circle $ABCD$ and the tangent $ST$ to the circle at point $C$.
The point $B$ lies on $AT$ such that $AB = BT$. The point $C$ lies on the line $ST$ such that $SC = CT$.

1. Show that $BC$ is parallel to $AS$.

In $\Delta AST$, we are given that $AB = BT$ and $SC = CT$.
By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side.
Therefore, $BC \parallel AS$.

2. Prove that $\angle BCA = \angle SCD$.

Since $BC \parallel AS$ and $A, D, S$ are collinear, $BC \parallel AD$.
$\angle BCA = \angle CAD$ (Alternate angles, $BC \parallel AD$)
$ST$ is a tangent to the circle at $C$.
By the Alternate Segment Theorem, the angle between the tangent $ST$ and chord $CD$ is equal to the angle in the alternate segment:
$\angle SCD = \angle CAD$
Therefore, $\angle BCA = \angle SCD$.

3. Prove that $\angle ABC = \angle SDC$.

$ABCD$ is a cyclic quadrilateral.
$\angle ABC + \angle ADC = 180^\circ$ (Angles in opposite segments are supplementary)
Since $A, D, S$ are collinear, $\angle SDC + \angle ADC = 180^\circ$ (Angles on a straight line)
Therefore, $\angle ABC = \angle SDC$.
(Alternatively, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle).

4. Conclude that the triangles are similar.

In $\Delta ABC$ and $\Delta SDC$:

1. $\angle BCA = \angle SCD$
2. $\angle ABC = \angle SDC$
   Since two pairs of corresponding angles are equal, the third pair must also be equal.
   By AA (Angle-Angle) similarity, $\Delta ABC$ is similar to $\Delta SDC$.

1. Use the similarity of $\Delta ABC$ and $\Delta SDC$ from part (i) to establish a ratio of corresponding sides.

Since $\Delta ABC$ is similar to $\Delta SDC$, the ratio of their corresponding sides is equal:
$\frac{AC}{SC} = \frac{BC}{DC}$
Cross-multiplying gives:
$AC \times DC = BC \times SC$

2. Apply the Midpoint Theorem to $\Delta AST$ to relate $BC$ and $AS$.

In $\Delta AST$, we are given that $AB = BT$ and $SC = CT$. This means $B$ and $C$ are the midpoints of $AT$ and $ST$ respectively.
By the Midpoint Theorem,
$BC = \frac{1}{2} AS$

3. Substitute the expression for $BC$ into the equation from step 1 to complete the proof.

Substituting $BC = \frac{1}{2} AS$ into $AC \times DC = BC \times SC$:
$AC \times CD = \left(\frac{1}{2} AS\right) \times SC$
$AC \times CD = \frac{1}{2} AS \times SC \text{ (shown)}$