On the side $BC$ of the rectangle $ABCD$, a point $P{}$ is marked so that $\angle APD = 90^\circ$. On the straight line $AD$, points $Q{}$ and $R{}$ are selected outside the segment $AD$ such that $AQ = BP$ and $CP = DR$. The circle $\omega$ passes through the points $Q, D$ and the circumcenter of the triangle $PDQ$. The circle $\gamma$ passes through the points $A, R$ and the circumcenter of the triangle $APR$. Prove that the radius of one of the circles touching the line $AD$ and the circles $\omega$ and $\gamma$ is $2AB$.