According to the figure, what is the area of the triangle $BCD$ of type $x$? 
$Q:$ 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.
 A: 
First of all note that $\angle ABC=\alpha+\beta=\angle ACB$, that means triangle ABC isosceles, so :
$\angle ABC=\angle ACB=45^o\Rightarrow \alpha=\beta=22.5^o$
Triangle AEC is right angled and we have:
$AC^2=r^2+x^2\Rightarrow AC=\sqrt{r^2+x^2}$
in triangle ABC:
$BC=AC\sqrt 2=\sqrt{2(r^2+x^2)}$
$AH=AC \sin 45^o=\frac{\sqrt {(r^2+x^2)}}{\sqrt 2}=HC$
$DH=HC tan 22.5^o=\frac{BC}2 tan 22.5^o=\frac{\sqrt{2(r^2+x^2)}}2 tan 22.5$
Area of DBC is:
$S_{DBC}= DH\times HC=\frac{\sqrt{2(r^2+x^2)}} 2 tan 22.5^o\times \frac{\sqrt {(r^2+x^2)}}{\sqrt 2}=\frac12(r^2+x^2)tan 22.5^o$
A: Reflect $A$ about $BC$ to $A'$, to form a a square $A'BAC$. Draw the circle of center $A'$ and radius $r=A'B=A'C$, intersecting circle $\omega$ at $D$. By construction $\angle BDC=135°$, which ensures $∠=∠=\alpha$ and $∠=∠=\beta=45°-\alpha$.
We have then
$$
BD= 2r\sin\alpha,\quad CD=2r\sin\beta
$$
and
$$
\text{area}_{BCD}={1\over2}BD\cdot CD\cdot\sin135°=
\sqrt2 r^2\sin\alpha\sin\beta=r^2\sin\alpha(\cos\alpha-\sin\alpha).
$$
On the other hand, from the cosine rule applied to $DAB$ one gets:
$$
AD^2=r^2+4r^2\sin^2\alpha-4r^2\sin\alpha\cos\alpha,
$$
whence:
$$
x^2=CE^2=r^2-AD^2=4r^2(\sin\alpha\cos\alpha-\sin^2\alpha)
=4\cdot\text{area}_{BCD}.
$$

