Let $ABCD$ be a convex quadrangle such that $AB=AC=BD$ (vertices are labelled in circular order). The lines $AC$ and $BD$ meet at point $O$, the circles $ABC$ and $ADO$ meet again at point $P$, and the lines $AP$ and $BC$ meet at the point $Q$. Show that the angles $COQ$ and $DOQ$ are equal.
Let $QO$ cut $(ADO)$ at $S$ . $Q$ lies on radical axis of $(ABC) , (ADO)$ $\implies$ $BDOS$ is cyclic .
So $\angle SBD = \angle SCA$ also we have $\angle SDB = \angle SAC$ So $\triangle SAC = \triangle SDB$ $\implies$ $SA=SD$ .
Hence $\angle DOQ =\angle DAS = \angle SDA = \angle SOA = \angle COQ$
$\blacksquare$
The problem is quite easy,just observe that $Q$ is the radical center of $ABC,AOD$ and $BCO$,so let $R$ be the intersection point of circumcirlcles of $BOD$ and $AOD$,so from here we are done.