The circle $\Omega$ centered at $O$ is the circumcircle of the triangle $ABC$. Point $D$ is chosen so that $BD \perp BC$ and points $A$ and $D$ lie in different half-planes with respect to the line $BC$. Let $E$ be a point such that $\angle ADB=\angle BDE$ and $\angle EBD+\angle ACB=90$. Point $P$ is chosen on the line $AD$ so that $OP \perp BC$. Let $Q$ be an arbitrary point on $\Omega$, and $R$ be a point on the line $BQ$ such that $PQ \parallel DR$. Prove that $\angle ARB=\angle BRE$. (All angles are oriented in the same way)
