IRL_ABooklet

(a) Because $OP$ is the perpendicular bisector of $BC$, $△PBC$ 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 $△QAC$ 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 $A^{\prime }$ be the second intersection point of the line $PM$ and the circumcircle of triangle $MQD$. Then looking at the cyclic quadrilateral $A^{\prime }MQD$ we see that $\angle P{A}^{\prime }D=\angle CQP=\angle A$ and $\angle QD{A}^{\prime }=\angle QMP=90^{\circ }$.

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

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



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