$\textbf{Problem}$: The points $D$ and $E$ are marked on the sides $AB$ and $AC$ of the triangle $ABC$, respectively. Points $B$, $C$, $E$, and $D$ lie on the same circle. Find the radius of the circle described near the triangle $ADC$, if it is known that $\angle CDE = \angle BAC$ and that the radius of the circle described near the triangle $ABC$ is $1$.
$\textbf{My solution attempts}$: We can apply the sine theorem to a triangle $ABC$ and get $\frac{BC}{\sin \angle A} = 2 R$. Since the quadrilateral $DBCE$ is inscribed in a circle we get $\angle B + \angle E = 180, \angle D + \angle C = 180$. I don't know where to go next.
It can be seen that triangles ADE and ABC are isosceles and $R_{ADC}= 1$ :
$\angle EDC+\angle EDA=\angle B+\angle BCD$
$\angle BDE+\angle BCE= \angle BDE + \angle ADE=180^o$
So: $ \widehat {BCE}=\widehat {EAD}$
So: $BC || DE$
$\widehat {EDC} = \widehat A=\widehat{DCB}$
Also : $\widehat{ADE}=\widehat{ABC}=\widehat B$
$(\widehat {ADE=B})+\widehat A+\widehat {BCD}=180^o$
That means $\widehat BDC=\widehat {BCD}$ and triangle $ABC$ is isosceles.
$\widehat{ADC}=\widehat A +\widehat B$
So:
$2R=\frac {AC}{Sin(A+B)}$
Due to statement we have:
$\frac {AC}{Sin B}=2\times 1=2$
also we have:
$\widehat {ADC}=\widehat A + \widehat B$
That means:
$2R=\frac{2 Sin (B)}{Sin (A+B)}$
$ \widehat A=180-2 \widehat B$
$Sin(A+B)=Sin(180-B)=Sin(B)$
$R=\frac {Sin B}{Sin B}=1$