[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.759399629840164, xmax = 36.138405689885204, ymin = -11.313040747702441, ymax = 25.48846375272708; /* image dimensions */ /* draw figures */ draw((3.6440275149831827,8.59384469174244)--(-5.262412736153678,0), linewidth(0.8) + blue); draw((-5.262412736153678,0)--(0,0), linewidth(0.8) + blue); draw((0,0)--(0,12.370421357932576), linewidth(0.8) + blue); draw((0,12.370421357932576)--(3.6440275149831827,8.59384469174244), linewidth(0.8) + blue); draw((3.6440275149831827,8.59384469174244)--(11.936248032194039,0), linewidth(0.8) + blue); draw((11.936248032194039,0)--(0,0), linewidth(0.8) + blue); draw((0,12.370421357932576)--(-5.262412736153678,0), linewidth(0.8) + blue); draw(circle((-2.631206368076839,6.185210678966286), 6.721612759941375), linewidth(0.8) + red); draw((-2.6269463811944673,6.195224684966264)--(11.936248032194039,0), linewidth(0.8) + blue); draw((-2.6269463811944673,6.195224684966264)--(0,0), linewidth(0.8) + blue); draw((-2.6269463811944673,6.195224684966264)--(3.6440275149831827,8.59384469174244), linewidth(0.8) + blue); /* dots and labels */ dot((0,0),dotstyle); label("$B$", (0.14530243894192235,0.38952972105338995), NE * labelscalefactor); dot((-5.262412736153678,0),dotstyle); label("$A$", (-5.0900580339225545,0.38952972105338995), NE * labelscalefactor); dot((0,5.077714157749999),dotstyle); label("$X$", (0.14530243894192235,5.470909003539474), NE * labelscalefactor); dot((3.6440275149831827,8.59384469174244),dotstyle); label("$C$", (3.802355710428138,8.973981084647304), NE * labelscalefactor); dot((0,12.370421357932576),dotstyle); label("$D$", (0.14530243894192235,12.746520248917276), NE * labelscalefactor); dot((11.936248032194039,0),linewidth(4pt) + dotstyle); label("$P$", (12.078844693265362,0.3125391258642069), NE * labelscalefactor); dot((-2.6269463811944673,6.195224684966264),linewidth(4pt) + dotstyle); label("$Q$", (-2.4723777974903163,6.510282038593446), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Quite quick exercise $\color{black}\rule{25cm}{1pt}$ The angle condition implies that $ABCD$ is cyclic, where $Q$ is the center of the circle. Because of this we have that $X$ is the orthocenter of $PAD$. Thus we have that $\angle PQC = \angle BDP = \angle CAP = \angle BQP$. Thus $PQ$ bisects $\angle BQC$.

parmenides51 wrote: Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^o$. Prove that $QP$ is the angle bisector of $\angle BQC$. Simple one, obviously, $X$ is the orthocenter of $\triangle ADP$. Since it is well known that the orthocenter is the incentre of the orthic triangle, the conclusion follows