[![figure](https://i.stack.imgur.com/wRSoJ.png)](https://i.stack.imgur.com/wRSoJ.png)
$Q:$ In a circle $\omega \ $ with a center $A$, $E\in \omega \ $ is the tangent point, $[BA]\bot[AC] \ $, $\angle ABD= \angle DCB \ $, $\angle DBC= \angle DCA \ $, $D\in\omega \ $, $|EC|=x \ $ What is the area of the triangle $BCD$ of type $x$?

$\text{Try on me:}$ I drew $[AD]$ and named it $\angle BAD=\theta$ and $|BD|=a, |DC|=b$. I also knew that $\alpha+ \beta =45^{\circ}$. I first found $\frac{\sin{\beta}}{\sin{\alpha}}=\frac{b}{a}$ by applying the sine theorem in the triangles $\triangle ABD$ and $\triangle DCA$. Then $\tan{\theta}=\frac{a^2}{b^2}$ came out of the trigonometric ceva theorem. I have drawn a right triangle between the point $D$ and the right part of $\triangle DCA$ by selecting the constant of the ratio $k$. By trying a little, I got the following two equivalences, $r$ is the half-diameter of the circle: $$2r^2+2x^2=\frac{1}{2k^2},$$ $$a^4+b^4+\sqrt{2}ab=\frac{3}{4k^2}.$$ I tried to make a few more calculations and make observations by choosing $r=1$, but I couldn't get anything. Can you help? Any help will be greatly appreciated.