Geometry Problem on triangle, tangents and a circle I have tried to solve this problem for very long all to no avail and none of my peers have been able to find a solution either. Can someone provide any clues?
"In the diagram B and D are points of contact of tangents drawn from A to
the circle. E is a point on the circle and C is the point on BE such that
AC ∥ DE."
"Prove ABCD is a cyclic quadrilateral".

What I know is that for it to be a cyclic quadrilateral, the opposite angles have to be supplementary and that the angle in the tangent is equal to the angle in the alternate segment. I already know that angleBED = angle BDA (alternate angle theorem) but still cannot make headway. But despite tons of angle labelling, I can't seem to get anything at all!
pls help
 A: 
The red lines are cut by a transversal BCE so we have $\alpha=\beta$ as equal alternate angles.
Angles in the alternate segment at D and E are equal so we have $\beta=\gamma$. Remove common angle $\beta$
So $ \alpha=\gamma$ are seen as equal angles on same side of AB, and by virtue of the converse theorem that when equal angles are subtended on same side of AB at C and D, we should have quadrilateral ABCD concyclic inside the green circle.
A: Hint: The converse of "angles in the same segment" holds. That is, if two angles subtended from a single line segment are equal and on the same side of the line segment, then the four points they define form a cyclic quadrilateral.
This reduces the problem to showing any one of the below pairs are equal (some are easier than others).

A: In figure $1$, radius $\overline{OB}$ and $\overline{OD}$ determine a cyclic quadrilateral $ABOD$ because they are perpendicular to tangents $\overline{AB}$ and $\overline{AD}$ respectively so we have $\angle{BAD}=180^{\circ}-2t$.
Besides, because of $\overline{ED}$ is parallel to $\overline{CA}$ we have $\angle{BCA}=t$. Consequently, in order that $ABCD$ be a cyclic quadrilateral it is necessary that $\angle{BCA}=\angle{ACD}=t$ because $\angle{BCD}$ should be suplementary of $\angle{BAD}$.
On the other hand the corresponding circumcircle is easily constructed in figure $2$ since it is the circumcircle of the triangle $\triangle{ABO}$ of radius $\dfrac{\overline{OA}}{2}$ and center the midpoint of $\overline{OA}$. This circle shows that in fact $\angle{BCA}=\angle{ACD}$ because $\overline{AB}=\overline{AD}$.
NOTE.- Not only points $A,B,C,D$ are on the same circle but also the center of the original circle of the problem. We could say that the irregular pentagon $ABCOD$ is cyclic and that if the first circle is centered at the origine the equation of the circle solution is $x^2-2rx+y^2=0$ where $2r=\overline{OA}$. Finally, on the red circle, point $C$ distinct of $A$ and $D$ can be any in the arc able to segment $\overline{BD}$ seen under the angle $2t$.

A: Since $\angle ECD$ is the supplement of $\angle DCB$, $ABCD$ is a cyclic quadrilateral if$$\angle ECD=\angle BAD$$

We start with a kite $HAGJ$, touching its inscribed circle at $B$, $D$, $E$, $F$, and complete the isosceles trapezoid $BDEF$.
Let kite diagonal $AJ$ meet trapezoid diagonal $BE$ at $C$. From the bilateral symmetry of the kite and isosceles trapezoid, it is clear that $C$ lies on the line of symmetry $HG$ and on the other trapezoid diagonal $DF$, and further that $AC\parallel DE$ since both are perpendicular to $HG$.
Therefore, by symmetry $\triangle CED$ is isosceles.
And since$$\angle CED=\angle ABD$$(tangent/alternate segment), then$$\triangle CED\sim\triangle ABD$$and$$\angle ECD=\angle BAD$$Conversely then: Given any triangle $BED$ inscribed in a circle, with tangents $AB$, $AD$, and with $AC$ drawn parallel to base $ED$, it is clear from the symmetry of the constructible kite and isosceles trapezoid that$$\angle ECD=\angle BAD$$ whence $\angle BAD$ is supplementary to $\angle BCD$, and $ABCD$ is a cyclic quadrilateral.
