In the diagram; the vertices A, B and C of triangle ABC are concyclic. EB and EC are tangents to the circle at B and C respectively. T is a point on AB such that TE is parallel to AC. BC cuts TE at F. Prove that B1 = T3. Prove that TBEC is a cyclic quadrilateral.

$\hat{B}_1 = \hat{A}$ (Angle between tangent EB and chord BC is equal to the angle in the alternate segment, $\angle BAC$).

We are given that $TE \parallel AC$.
$\hat{T}_3 = \hat{A}$ (Corresponding angles, with $TE \parallel AC$ and AB as the transversal).

Since both $\hat{B}_1$ and $\hat{T}_3$ are equal to $\hat{A}$, it follows that $\boxed{\hat{B}_1 = \hat{T}_3}$.

From 4.1, we have proven that $\hat{B}_1 = \hat{T}_3$.

In quadrilateral TBEC, $\hat{B}_1$ is $\angle EBC$.
In quadrilateral TBEC, $\hat{T}_3$ is $\angle ETC$.

Since $\angle EBC = \angle ETC$, these are angles subtended by the same chord EC at points B and T. If two angles subtended by the same chord (or arc) at the circumference are equal, then the four points are concyclic.

Therefore, $\boxed{\text{TBEC is a cyclic quadrilateral}}$.