Let $X =DQ\cap CP$ and $Y= AB\cap BC$. Then line $XY$ is polar to $M$, so $XY\bot OM$. Since $OM\bot PQ$ we have $XY||PQ$. Let $S=DC\cap XY$. Now: $(Q,P;M,\infty)=(XQ,XP;XM,XY)=(XD,XC;XM,XY)=(D,C;M,S) =(YD,YC;YM,YS)= (YA,YB;YM,YS)= (A,B;M,\infty)=-1$ So $M$ on the middle of segment $PQ$ and conclusion folows.

Zhaoli could you post a solution with a picture of the problem please

Why? A solution has been posted already, and it's not hard to draw the picture yourself.

Please maybe zaoli has another solution

silouan wrote: Please maybe zaoli has another solution it is easy to draw a figure by yourself! Solution 1 (From zhaobin) Since $D, O, M, Q$ are concyclic, So $DQ = \frac{OD}{\sin \angle OMD} \cdot \sin \angle DMQ.$ (sine law) Similarly, $CP = \frac{OC}{\sin \angle OMC} \cdot \sin \angle CMP.$ While $\angle OMD + \angle OMC = \pi$, $\angle DMQ = \angle CMP$, $OC = OD$, so $CP = DQ.$ $PB \cdot PA = CP^2 = DQ^2 = QA \cdot QB.$ While $PB \cdot PA = (PM - MB)(PM + MB) = PM^2 - MB^2, QB \cdot QA = (QM + MA)(QM - MA) = QM^2 - MA^2.$ So $PM = QM$, which means $QA = PB$. Solution 2 (From lin) Since $CD$ passes through $M$, so $C, D$ lie on the opposite sides of the diameter through $OM$. So $P, Q$ lie on opposite direction outside segment $AB$. Since $OM$ is perpendicular to $AB$, $OC$ is perpendicular to $PC$, so $O, M, C, P$ concyclic, $\angle OPM = \angle OCM.$ Similarly, $\angle OQM = \angle ODM$, triangles $OCD$ and $OPQ$ are similar. But $OC = OD$, $\angle OCM = \angle ODM$, so $\angle OPM = \angle OQM$, $OP = OQ$, result in $PM = QM$, $PA = QB$.

Since $\angle{QMO}=\angle{QDO}=90^o$, so $Q, D, M, O$ are cyclic. Hence $\angle{QOD}=\angle{QMD}$. Similar, we have $C, P, M O$ are cyclic. Therefore $\angle{COP}=\angle{CMP}$. Because $\angle{CMP}=\angle{DMQ}$, so $\angle{QOD}=\angle{COP}$. Along with $OD=OC$, $\angle{ODQ}=\angle{OCP}$, we will have: $\triangle{ODQ}\cong\triangle{OCP}$, which gives $OQ=OP$. Therefore $\triangle{OPB} \cong \triangle{OQA}$, hence $QA=PB$. P.S. Hope this picture will help you, silouan.

Attachments:

thank you very much.