In the diagram below, $TAX$ and $TCY$ are tangents to the circle at $A$ and $C$ respectively. $AC$ meets $TD$ at $E$ and $D$ is on $BC$ such that $TD$ is parallel to $AB$.

1. Relate angle ACB to the angle between the tangent and the chord.

$\angle TAB = \angle ACB$ (Alternate segment theorem)

2. Relate angle TAB to angle ATD using parallel lines to complete the proof.

$\angle ATD = \angle TAB$ (Alternate angles, $TD \parallel AB$)

Therefore, $\angle ACB = \angle ATD$ (Shown)

1. Relate the angle from part (a) to the points in question

From (a), $\angle ATD = \angle ACB$.
Since $B$, $D$, and $C$ are collinear, $\angle ACB$ is the same angle as $\angle ACD$.
Therefore, $\angle ATD = \angle ACD$.

2. Conclude that the points are concyclic

Since the angles subtended by the line segment $AD$ at points $T$ and $C$ are equal ($\angle ATD = \angle ACD$), the points $T$, $A$, $D$, and $C$ lie on the same circle (converse of angles in the same segment).

1. Identify equal angles to prove triangles are similar

Consider $\triangle TEC$ and $\triangle AED$.
$\angle TEC = \angle AED$ (vertically opposite angles)
Since $T, A, D, C$ are concyclic (from part b), $\angle TCE = \angle ADE$ (angles in the same segment subtended by arc $AT$).

2. Conclude similarity and establish the ratio of corresponding sides

Therefore, $\triangle TEC$ is similar to $\triangle AED$.
By the ratio of corresponding sides,
$\frac{CE}{DE} = \frac{TE}{EA}$

3. Rearrange the ratio to obtain the required product

Cross-multiplying gives:
$CE \times EA = DE \times TE \text{ (Shown)}$