In the diagram, TE and TC are tangents to the circle at A and B respectively. BD is parallel to TE.

1. Identify equal angles using the alternate segment theorem

Let $\angle TAB = x$.
By the alternate segment theorem, $\angle ADB = \angle TAB = x$.

2. Identify equal angles using parallel lines

Since $TE \parallel BD$, alternate angles are equal.
Therefore, $\angle ABD = \angle TAB = x$.

3. Conclude that the triangle is isosceles

In $\triangle ABD$, $\angle ADB = \angle ABD = x$.
Since the base angles are equal, $\triangle ABD$ is an isosceles triangle.
Therefore, $AB = AD$.

1. Relate the angle between the tangent and chord to an angle in the alternate segment.

Let $\angle TAB = x$.
By the alternate segment theorem, the angle between the tangent $TA$ and chord $AB$ is equal to the angle in the alternate segment:
$\angle ADB = \angle TAB = x$.

2. Use parallel lines to find another equal angle.

Since $BD$ is parallel to $TE$ (which is the line $TA$), by alternate angles:
$\angle ABD = \angle TAB = x$.

3. Use tangent properties to find base angles of the isosceles triangle formed by the tangents.

Tangents from an external point $T$ are equal in length, so $TA = TB$.
Therefore, $\triangle TAB$ is an isosceles triangle, and its base angles are equal:
$\angle TBA = \angle TAB = x$.

4. Prove that triangle TAB is similar to triangle ABD and deduce the required relationship.

In $\triangle TAB$ and $\triangle ABD$:

1. $\angle TAB = \angle ABD = x$
2. $\angle TBA = \angle ADB = x$
   By AA similarity, $\triangle TAB$ is similar to $\triangle ABD$.

Using the ratio of corresponding sides:
$\frac{TA}{AB} = \frac{AB}{BD}$
Cross-multiplying gives:
$(AB)^2 = TA \times BD$