IRL ABooklet

(a) Because $OP$ is the perpendicular bisector of $BC$, $\triangle PBC$ is isosceles with symmetry axis $PO$. Thus $|PB| = |PC|$ and $\angle PBC = \angle PCB = \angle C$. Hence $\angle APB = 2\angle C$ (external angle).
Similarly, $OQ$ is the perpendicular bisector of $AC$ hence $\triangle QAC$ is isosceles with symmetry axis $QO$. Thus $|QA| = |QC|$ and $\angle QAC = \angle QCA = \angle C$. Hence $\angle AQB = 2\angle C$ (external angle).

This implies that quadrilateral $APQB$ is cyclic with $\angle APB = \angle AQB$. Because $\angle AOB = 2\angle C$ by the Central Angle Theorem, the point $O$ lies on this circle as well, i.e. $A, P, O, Q, B$ are all on the same circle.
To show that $J$ is on this circle too, we consider the parallelogram $CPJQ$.
Because $JP \parallel BC$ and $PO \perp BC$, we have $\angle OPJ = 90^\circ$. Similarly, using $JQ \parallel AC$ and $QO \perp AC$ we obtain $\angle OQJ = 90^\circ$. This implies that the four points $P, O, Q, J$ are on the circle with diameter $OJ$. Hence, $A, B, O, P, Q, J$ are all on the same circle.

(b) The quadrilateral $CNOM$ is cyclic because $\angle CNO = 90^\circ$ and $\angle CMO = 90^\circ$, and $OC$ is a diameter of its circumcircle. The quadrilateral $MNPQ$ is cyclic as well since $\angle PNQ = \angle PMQ = 90^\circ$. The line $MN$ is the radical axis of the circumcircles of $CNOM$ and $MNPQ$.

On the other hand, $PQ$ is the radical axis of the circumcircles of $APOQB$ and $MNPQ$. Hence the point $D$ is the radical centre of the three circles mentioned above. As the circumcircles of $ABPOQ$ and $CMON$ have the point $O$ in common, it follows that $OD$ is the radical axis of these two circles and so is perpendicular to the line that connects the centres of them. These circles have diameters $OJ$ and $OC$, hence the centre line connects the midpoints of $OJ$ and $OC$ and therefore is parallel to $CJ$. Hence $OD \perp CJ$.

Alternative proof of (b).

Using the short notation $\angle BAC = \angle A$, $\angle ABC = \angle B$, $\angle BCA = \angle C$, we have $\angle CNM = \angle A$ since $NM \parallel AB$. Because $CNOM$ is cyclic due to the right angles at $M$ and $N$, we have $\angle MOQ = \angle C$. Also, $\angle CQP = \angle A$ and $\angle CPQ = \angle B$ as $APQB$ is cyclic by part (a).

Let $A'$ be the second intersection point of the line $PM$ and the circumcircle of triangle $MQD$. Then looking at the cyclic quadrilateral $A'MQD$ we see that $\angle PA'D = \angle CQP = \angle A$ and $\angle QDA' = \angle QMP = 90^\circ$.

Extend $A'D$ to intersect $QN$ at $B'$. Note that triangles $OA'B'$ and $CQP$ are similar, in particular $\angle A'B'O = \angle B$.

Since $PNDB'$ is cyclic due to the right angles at $N$ and $D$, we have $\angle PB'N = \angle PDN = \angle CNM - \angle CPQ = \angle A - \angle B$ where the last equality comes from an external angle of triangle $DNP$. Therefore $\angle PB'D = \angle A'B'O + \angle PB'N = \angle B + \angle A - \angle B = \angle A$. Thus the triangle $A'PB'$ is isosceles, and because $PD$ is perpendicular to $A'B'$, $D$ is the midpoint of $A'B'$.

Let $D'$ be the intersection of $CJ$ and $PQ$, and note that $D'$ is the midpoint of $PQ$ since $PCQJ$ is a parallelogram.
From the angle equalities established above we see that triangle $QPC$ (with $D'$ as midpoint of $QP$) is similar to triangle $A'B'O$ (with $D$ as midpoint of $A'B'$). Thus $\angle QCD' = \angle A'OD$, and so, since $OA' \perp CQ$, $OD$ is perpendicular to $CD'$, i.e. to $CJ$.