$Q:$ 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.
Best Answer
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 $\angle BDC=135Β°$, which ensures $β π΄π΅π·=β π·πΆπ΅=\alpha$ and $β π·π΅πΆ=β π·πΆπ΄=\beta=45Β°-\alpha$.
We have then $$ BD= 2r\sin\alpha,\quad CD=2r\sin\beta $$ and $$ \text{area}_{BCD}={1\over2}BD\cdot CD\cdot\sin135Β°= \sqrt2 r^2\sin\alpha\sin\beta=r^2\sin\alpha(\cos\alpha-\sin\alpha). $$ 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: $$ x^2=CE^2=r^2-AD^2=4r^2(\sin\alpha\cos\alpha-\sin^2\alpha) =4\cdot\text{area}_{BCD}. $$