Here's my solution: Let $PQ \cap RS=M$, and let $O$ be the center of $\omega_2$. Note that $\angle OPX=\angle OQX=90^{\circ}$, and so $P,O,Q,X$ are concyclic. Then $O$ is the center of spiral similarity that takes $\overline{PQ}$ to $\overline{RS}$, which gives $\triangle OPQ \sim \triangle ORS$. But, as $OP=OQ$, so we get $OR=OS$. Also, $\angle POQ=\angle ROS$, which gives that $\overline{OQ}$ and $\overline{OR}$ are isogonal in $\angle POS$. Then, by the Isogonal Line Lemma, we get that $OX$ and $OM$ are also isogonal in $\angle POS$ (as $\overline{PR} \cap \overline{QS}=X$ and $\overline{PQ} \cap \overline{RS}=M$). This gives $\angle POM=\angle SOX=\angle SRX=\angle MRX$, which means that $POMR$ is also cyclic. Thus, $\angle OMR=\angle OPM=90^{\circ}$, which implies $\overline{OM} \perp \overline{RS}$. But, as $OR=OS$, so we get that $M$ is the midpoint of $\overline{RS}$. Hence, done. $\blacksquare$

I have the same solution as above, but after showing $OR = OS$, can't we just note that $OPMR, OPMS$ are concyclic (due to the spiral similarity center), so $\angle OMR = \angle OPR = 90$, so $OM \perp RM \Rightarrow OM \perp RS$, and thus with with $OR = OS$, we get $RM = MS$ ?

Here's my solution, with a bit of trig to simplify the end: Let $PQ \cap RS=N$, and consider $O$ the center of $\omega_2$. As $XP, XQ$ are tangents, then $\angle OPX = \angle OQX = 90º, OX$ is common hypotenuse, and $OP = OQ$, so $\triangle OQX \equiv \triangle OPX \implies \angle XOS = \angle XOR$, as $X, O, R, S \in \omega_2$, then $OR = OS$, too $\angle OQS = \angle OPR = 90º$ and $OP = OQ \implies \triangle OPR \equiv \triangle OQS.$ Then, $PR = QS.$ By law of sines at $\triangle NSQ \implies \frac{sin(SNQ)}{SQ} = \frac{sin(NQS)}{NS}$ at $\triangle NPR \implies \frac{sin(PNR)}{PR} = \frac{sin(RPN)}{NR}.$ Using that $SQ = PR$ and $\angle SNQ = \angle PNR \implies \frac{sin(NQS)}{sin(NPR)} = \frac{NS}{NR}$, but $\angle NPX = \angle NQX = \angle NQS.$ Thus, $\angle NPR = 180º - \angle NQS \implies sin(NPR) = sin(NQS)$, replacing it at the last equation, we are done.

Redacted (Dumbness knows no bounds )

math_pi_rate wrote: Kayak wrote: I have the same solution as above, but after showing $OR = OS$, can't we just note that $OPMR, OPMS$ are concyclic (due to the spiral similarity center). For that wouldn't you have to show that $O$ is also the center of spiral similarity that takes $PM$ to $XS$? Or maybe I am missing something absolutely trivial here. $\angle ORP = \angle ORX = \angle OSX = \angle OQS$ and $\angle OPR = 90 = \angle OQS$, we have $O$ is the center of spiral similarity mapping $PR \mapsto QS$. If we let $PQ \cap RS = Y$, then $ \omega_{PYR} \cap \omega_{QYS}$ is the center of similarity mapping $PR \mapsto QS$, so $ \omega_{PYR} \cap \omega_{QYS} = O$ . Well most probably I am reasoning circularly/messing up how spiral centers are defined

Am I missing sth here? Obviously $\angle PXO=\angle OXQ$ from the congruency of the two triangles. Equivalently we can rewrite it as $\angle RXO=\angle OXS$ implying that the chords $RO, OS$ are equal. Now since $PQ$ is the simson line of$ \triangle XRS$ wrt $O$, we can imply that $M$ is the foot of $O$ onto $RS$. However, since $\triangle ORS$ is isosceles, $M$ is also midpoint. Hence we are done.

let $O_2 $be $\omega_2$ circumcenter let T be the intersection of the PQ and RS $O_2PXS $ and $O_2QXP $ are cyclc then $\angle RO_2S=180-\angle PXQ=\angle PO_2Q$ so $O_2P$ and $O_2P$ are two isogonal in the triangle $ RO_2Q$ (i) let $T' \ in \ RS$ be the feet of the altitudes from $O_2$ so $O_2T'SQ $ and $O_2RPT'$ are cyclc by chase angle and using (i) we ll show that T' P Q are colliniar therefore T'=T $\frac{O_2S}{O_2R}=\frac{sin(PRO_2)}{sin(O_2SQ)}=\frac{sin(O_2TP)}{sin(O_2TQ)}=1$ then $O_2T $ is the mediatrice of $RC$

Though maybe same as above but still posting for practice . Let $T$ lie on $RS$ and $OT$ is perpendicular to $RS$ .Since $O$ lies on $(XRS)$ we conclude that $P,T,Q$ are collinear. Now if $OR=OS$ we will be done . But its clear that $\angle PRO = \angle XRO =\pi - \angle OSX = \angle OSQ $. Also $OP=OQ$ via the tangency condition .Therefore $OPR \sim OQS \implies OR=OS$ so we are done.

Let O1 and O2 be the centers of the circles, M the intersection of PQ and RS and M' be the midpoint of RS. Now,quadrilaterals XPO2Q and XRO2S are cyclic. So,<PO2Q=180 - <X=<RO2S. Again <O2RS=<O2XS=<O2XQ=<O2PQ. So, triangles O2RS and O2PQ are similar. As, O2P=O2Q so, O2R=O2S. So,O1,M'and O2 are collinear and <O2M'R=90. Now, <O2M'R+<O2PR=90+90=180. So, O2PRM' is cyclic. Again,<O2PM=<O2PQ=<O2RS=<O2RM. So, O2PRM is cyclic. So, (O2PR) is the circumcircle of triangle RMM'.But points R,M,M' are collinear. So, M'=M and PQ passes through the midpoint of RS.