Show that the quadrilateral $BCED$ is inscribed...

Question

Problem

The triangle $ABC$, with $AB>AC$ and $\angle A \neq 90$, is inscribed with a center circle $O$, and $T$ is the diametrical point opposite $A$. The tangent in $T$ to the circle intersects the lines $AB$ and $AC$ in $D$ and $E$ respectively. Show that the quadrilateral $BCED$ is inscribed.

The drawing:

MY IDEA:

So, to show that $BCED$ is inscribed we can show that it is an isosceles trapezium or we can show that opposite angles are supplementary.

As the quadrilateral doesn't look like an isosceles trapezium, I thought of the second one.

We can clearly see that

$\angle ACT= \angle ECT= \angle ABT= \angle ATE= \angle ATD= 90$

Also, we can see that

$\angle ETC= \angle CAT$

$\angle CTA= \angle CET$

$\angle TAD= \angle BTD$

$\angle ATB= \angle BDT$

I don't know what to do forward. I hope one of you can help me! Thank you!

Answer 1

Following Reza Rajaei's comment, note that $\angle ATB = \angle BDE$. Then, using the fact that $ABTC$ is a cyclic quadrilateral, we get that $\angle ACB = \angle ATB$. Since $\angle ECB = \pi - \angle BDE $, then the quatrilateral $ECBD$ is cyclic.

Answer 2

According to the markings on you figure, $\angle 1 = \angle 2$.

Will $\angle 2 = \angle 3$?