Let $ABC$ be acute-angled triangle . A circle $\omega_1(O_1,R_1)$ passes through points $B$ and $C$ and meets the sides $AB$ and $AC$ at points $D$ and $E$ ,respectively .
Let $\omega_2(O_2,R_2)$ be the circumcircle of triangle $ADE$ . Prove that $O_1O_2$ is equal to the circumradius of triangle $ABC$ .
Assume that $O$ is the circumcenter of triangle $ABC$ .
we Know $AO$ is perpendicular to $DE$ and $O_1O_2$ is perpendicular bisector of $DE$
alse we know $AO_2$ is perpendicular to $BC$ and $OO_1$ is perpendicular bisector of $BC$
thus $AOO_1O_2$ is a parallelogram and $AO=O_1O_2=R$
QED