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 an angle involving the tangent using the Alternate Segment Theorem.

$\angle TAB = \angle ACB$ (Alternate Segment Theorem)

2. Relate angle TAB to angle ATD using parallel lines.

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

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

1. Relate the angle from part (a) to the points T, A, D, C

From part (a), we have $\angle ACB = \angle ATD$.
Since D lies on the line segment BC, $\angle ACB$ is the same angle as $\angle ACD$.
Therefore, $\angle ACD = \angle ATD$.

2. Apply the converse of the angles in the same segment theorem to conclude concyclicity

Both $\angle ACD$ and $\angle ATD$ are subtended by the same line segment AD at points C and T respectively.
Since these angles are equal, by the converse of angles in the same segment, the points T, A, D, and C are concyclic.
Thus, a circle can be drawn passing through them.

1. State the concyclic points from part (b) and identify equal angles in the resulting circle.

From part (b), $T, A, D, C$ lie on a circle.
Consider $\triangle CDE$ and $\triangle TAE$:
$\angle CDE = \angle TAE$ (angles in the same segment subtended by arc $TC$)
$\angle DCE = \angle ATE$ (angles in the same segment subtended by arc $AD$)

2. Conclude that the triangles are similar.

$\angle CED = \angle TEA$ (vertically opposite angles)
Therefore, $\triangle CDE$ is similar to $\triangle TAE$.

3. Use the ratio of corresponding sides to prove the required equation.

Since the triangles are similar, the ratio of their corresponding sides is equal:
$\frac{CE}{TE} = \frac{DE}{EA}$
Cross-multiplying gives:
$CE \times EA = DE \times TE$