$M \in \gamma_1$ where $\gamma_1$ is the circumference circumscripted to OPQ which is; using chord-angles, we note that $\angle PMQ$ and $\angle QMO$ are fixed: so also $\angle QMO+0.5\angle PMQ$ is fixed: now prove the case of $M\equiv O$ m=o, and see that it's $\frac{\pi}{2}$

MP, MQ, MO and the bisector line of PMQ are an armonic fascicul. The problem becomes a known theorem

Bravo, Flavian ! It is and my proof.

Let $MH$, with $H$ on the circumscribed circle of $POMQ$, be the angle bisector of $\measuredangle PMQ$. And $PQ \cap MH = K$. Connect $O$ with $H$. It is $OH \bot PQ$. So $\measuredangle OPM + \measuredangle MKQ = 90^{\circ}$. $\measuredangle MKQ = 180^{\circ} - (\measuredangle KQM + \measuredangle KMQ) = 180^{\circ} - (180^{\circ} - \measuredangle POM + \measuredangle KMQ) = \measuredangle POM - \measuredangle KMQ = \\ = 180^{\circ} - \measuredangle OPM - \measuredangle PMO - \measuredangle KMQ$. Then $90^{\circ} - \measuredangle OPM = 180^{\circ} - \measuredangle OPM - \measuredangle PMO - \measuredangle KMQ$. And finally $90^{\circ} = \measuredangle PMO + \measuredangle KMQ = \measuredangle PMO + \measuredangle PMK$.

let $\omega$ be the circumcircle of $PMOQ$,given $OP=OQ$,Let $OL$ is the bisector of $O$($L$ on circumcircle),then $OL$ is the diameter,also bisector of $\angle PMQ$ meet at $L$,hence done.

Dear Mathlinkers, I saw this terminology: O, L is the first, second M-perpoint of triangle MPQ.... Sincerely Jean-Louis

That's really easy. Just assume $\angle\{POQ}=2\theta$ and proceed. Oil is the food of the hair; Rice is the food of the Bengalis; Maths is the food of my soul.

Let $ \omega $ denote the circumcircle of $ POMQ $, and let $ S $ be the intersection of the angle bisector of $ PMQ $ and $ \omega $. Let $ \angle SMP = \angle SMQ = \alpha $. Note that $ \angle POQ = \angle PMQ = 2\alpha $. Let $ T $ be the point diametrically opposite $ Q $. Then $ \angle POT = 180 - 2\alpha $. Since $ POT $ is isosceles, $ \angle PTQ = \alpha $, so $ \angle PQT = 90 - \angle PTQ = 90 - \alpha $. Thus $ \angle PMO = \angle PQO = \angle PQT = 90 - \alpha $, so $ \angle PMO + \angle SMP = (90 - \alpha) + \alpha = 90 $ and we are done.

k2c901_1 wrote: $P,Q$ are two fixed points on a circle centered at $O$, and $M$ is an interior point of the circle that differs from $O$. $M,P,Q,O$ are concyclic. Prove that the bisector of $\angle PMQ$ is perpendicular to line $OM$. I would like to present the following solution which basically trivializes the entire problem. In my opinion it is very fast and elegant . Solution: Denote by $C_1$ the circle with centre $O$ and by $C_2$ the other circle .Drop the perpendicular from point $O$ to segment $\bar{PQ}$ and extend it to the point diametrically opposite $O$, naming them $A$ and $X$ respectively. Clearly, since $OA \bot AI$, it suffices to prove that $O,A,I,M$ are concyclic. Now for the best part, consider an inversion with respect to $C_1$ ($O$ is the centre of inversion), This it suffices to prove that $A^*,I^*,M^*$ are collinear but this clearly follows from the fact that both $OX^*$ and $OM^*$ are diameters and thus $\measuredangle{OI^*X^*}= \measuredangle{M^*I^*O}= 90^o $. Thus done . Sincerely, Aayam

Dear Mathlinkers, here Problem 7, p. 20 Sincerely Jean-Louis