Claim: $\measuredangle DQB = \measuredangle APC$. Proof. Note that \[ \measuredangle (DQ, QB) = \measuredangle (DQ, DE) + \measuredangle (EB, BQ) = \measuredangle DAE + \measuredangle ECB = \widehat{DC} + \widehat{AB} \]which implies the result. $\blacksquare$ Define $C' = CP \cap (ABC)$ and $A' = AP \cap (ABC)$. Claim: We have that $\triangle DQB \cong \triangle A'PC$. Proof. We've shown one angle relation so now we prove the other to get similarity (modulo boring config issues). Note that \[ \measuredangle PC'A' = \measuredangle CAA' = \measuredangle EDA = \measuredangle BCE = \measuredangle QBE. \]Now, note that since $\measuredangle C'CA = \measuredangle CBE = \measuredangle CC'A$ we get that $CD' \parallel AC$. By symmetry, it thus follows that $C'D \parallel AC \parallel BA'$, and thus $\widehat{BD} = \widehat{C'A'}$. $\blacksquare$ Then rotating $BD$ to $C'A'$ implies the result.

complex moment Let $AP$ and $CP$ intersect $(ABCD) = \omega$ again at points $R$ and $S,$ respectively; then $$\measuredangle RAC = \measuredangle RAE = \measuredangle ADE = \measuredangle ADB,$$so $ABRC$ is an isosceles trapezoid. Similarly, $ADSC$ is an isosceles trapezoid. We now use complex numbers with $\omega$ as the unit circle. Let lowercase letters denote the complex numbers representing their corresponding uppercase letters, as usual. Then $r = \frac{ac}{b}$ while $s = \frac{ac}{d}.$ Thus \begin{align*} p &= \frac{ar(c+s)-cs(a+r)}{ar-cs} \\ &= \frac{\frac{a^2 c}{b} \left(c+\frac{ac}{d}\right) - \frac{ac^2}{d} \left(a + \frac{ac}{b}\right)}{\frac{a^2 c}{b} - \frac{ac^2}{d}} \\ &= \frac{\frac{a^2 c^2}{bd} (d+a) - \frac{a^2 c^2}{bd} (b+c)}{\frac{a^2 c^2}{bd} \left(\frac{d}{c} - \frac{b}{a}\right)} \\ &= \frac{ac(a+d-b-c)}{ad-bc}. \end{align*}Similarly, \[ q = \frac{bd(a+d-b-c)}{ad-bc}. \]In particular, $p = \frac{ac}{bd} q.$ Taking magnitudes gives the result.

Let $T = AD\cap BC$. We claim $OP=OQ=OT$. Let the tangents from $A$ and $C$ meets $\odot(ABCD)$ again at $A_1$ and $C_1$. Notice that $\measuredangle A_1AC = \measuredangle ADB$, so $CA_1=AB$. Thus, $B$ and $A_1$ are symmetric across $AC$. Hence, lines $BC$ and $AA_1$ are symmetric across perpendicular bisector $\ell_{AC}$ of $AC$. Similarly, lines $AD$ and $CC_1$ are symmetric across $\ell_{AC}$. Hence, $P$ and $T$ also are. Since $O\in\ell_{AC}$, this means $OP=OT$.

Let $AD$ and $BC$ intersect at $Z$. Claim: $ACPZ$ is concyclic $$\measuredangle (AP;CP)=\measuredangle(AO_1,CO_2)=\measuredangle(AD;CB)$$Claim: $ACPZ$ is isosceles trapezoid $$\angle PAC=\angle ADE=\angle ACZ$$ Now since $OA=OC$ we must also have $OP=OZ$. Similarly, $OQ=OZ$, finishing the problem.

$OP=OQ$ again? (see TSTST 2021/1) Interesting, well it didn't last longer than 10 minutes sadly. Claim 1: Let $BC \cap AD=F$ then $FBDQ, FACP$ are ISLsusceles trapezoids. Proof: We only need to prove one since the other is analogous, now note that $\angle FBQ=\angle BEC=\angle AED=\angle FDQ$ so $FBDQ$ is cyclic, and since $\angle FDB=\angle ACB=\angle DBQ$ we have its an ISLsusceles trapezoid, thus the claim is proven. Finish: On perpendicular bisector of $BD$ we get $OF=OQ$ but in perpendicular bisector of $AC$ we get $OF=OP$ therefore $OP=OQ$ as desired thus we are done .

Let the tangent to $\omega_1$ at $A$ intersect $BC$ at $X$, and let the tangent to $\omega_2$ at $C$ intersect $AD$ at $Y$. Notice that \[\angle YAC=\angle DAC=\angle DBC=\angle ECY,\]by tangential properties. This gives us that $\triangle AYC$ is isosceles with $AY=YC$. Similarly, we find that $\triangle AXC$ is isosceles with $AX=XC$. However, note that $AO=OC$, meaning that $X$, $O$, and $Y$ all lie on the perpendicular bisector of $AC$. Reflecting the figure across $XY$, we send line $AX$ to line $CX$, or line $BC$, and line $CY$ to line $AY$, or line $AD$. This means that $P$ is sent to \[P=AX\cap CY \implies P'=AD\cap BC.\]However, since $O$ is also on $XY$, the reflection of $O$ across $XY$ is just $O$, meaning that $OP=OP'$. Similarly, we can get that the reflection of $Q$ across $BD$ is also the intersection of $AD$ and $BC$, and that $OQ=OQ'$, since $O$ is on the perpendicular bisector of $BD$ (because $O$ is the center of $(ABCD)$). Now, since $P'=Q'$, this means that we must have that $OP=OQ$, as desired. This finishes the problem.

We use moving points. If $AE = DE$ the problem is true by symmetry, so we assume $AE \neq DE$. Fix $A$, $D$, and $E$, and move $B$ linearly along line $DE$. By e.g. Reim's theorem, $C$ moves linearly as well. How do we compute $O$? It lies on the perpendicular bisector of $AD$, which is fixed. It also lies on the perpendicular bisector of $BC$, which is parallel to a fixed line, and passes through the midpoint of $BC$, which moves linearly. Therefore, $O$ moves linearly as well. $P$ is the intersection of the tangent to $(ADE)$ at $A$, which is fixed, and the tangent to $(BCE)$ at $C$, which moves linearly, and therefore is linear. Similarly, $Q$ moves linearly. Therefore, to verify that $\overrightarrow{OP}\times \overrightarrow{OP} = \overrightarrow{OQ} \times \overrightarrow{OQ}$, it suffices to check three cases. The two cases where $EA = EB$ are trivial by symmetry. We will also check that $\lim_{B \to \infty_{DE}} \frac{\overrightarrow{OP}\times \overrightarrow{OP}}{\overrightarrow{OQ}\times \overrightarrow{OQ}} = 1$. Indeed, obviously $\lim_{B \to \infty_{DE}} \frac{|OP|}{|AP|} = \lim_{B \to \infty_{DE}} \frac{|OQ|}{|DQ|} = 1$. In addition, \begin{align*} \frac{dAP}{dDQ} &= \frac{dAP}{dAC}\times\frac{dAC}{dBD}\times\frac{dBD}{dBQ}\\ &= \frac{\sin \angle ACP}{\sin \angle APC}\times\frac{dEC}{dEB}\times\frac{\sin \angle BQD}{\sin \angle BDQ}\\ &= \frac{\sin \angle ACP}{\sin (\angle ACP + \angle CAP)}\times \frac{\sin \angle EBC}{\sin \angle ECB}\times\frac{\sin (\angle BDQ + \angle DBQ)}{\sin \angle BDQ}\\ &= \frac{\sin\angle BCE}{\sin(\angle BCE + \angle ADE)}\times \frac{\sin\angle DAE}{\sin\angle ADE}\times\frac{\sin(\angle DAE + \angle ECB)}{\sin\angle EAD}\\ &=\frac{\sin(\angle DAE + \angle ECB)}{\sin(\angle BCE + \angle ADE)}\\ &= 1. \end{align*} It follows that $\lim_{B \to \infty_{DE}}\frac{|AP|}{|DQ|} = 1$. Therefore, \begin{align*} \lim_{B \to \infty_{DE}}\frac{|OP|}{|OQ|} &= \lim_{B \to \infty_{DE}}\frac{|OP|}{|AP|}\times\lim_{B \to \infty_{DE}}\frac{|AP|}{|DQ|}\times\lim_{B \to \infty_{DE}} \frac{|DQ|}{|OQ|}\\ &= 1, \end{align*} which completes the $B \to \infty_{DE}$ case, and we are done.

First, notice that $$\angle AED=\angle BEC$$because they subtend the same arc $AEBC$ in the circumcircle of quadrilateral $ABCD$. This implies that $$\triangle AED \sim \triangle BEC$$by the angle-angle similarity criterion since $$\angle DAE=\angle EBC$$(both subtend arc $DE$) and $$\angle DEA=\angle ECB$$(both subtend arc $DC$). The tangents at $A$ and $D$ to $\omega_1$ and the tangents at $B$ and $C$ to $\omega_2$ form two kites $APDQ$ and $BPCQ$ because in each kite, one pair of adjacent sides are radii of $\omega_1$ and $\omega_2$ and the other pair are equal tangents from a common point to a circle. Therefore, $$\angle AP=\angle PD$$and $$\angle BQ=\angle QC$$. By the Radical Axis Theorem, the lines $AC$, $BD$, and the line joining the points of tangency of the external tangents (which is $PQ$) are concurrent. This implies that $PQ$ passes through the radical center $E$ of $\omega_1$, $\omega_2$, and the circumcircle of $ABCD$. The power of point $P$ with respect to $\omega_1$ equals the power of point $P$ with respect to $\omega_2$ because $$PA^2=PC^2$$(tangents from a point to a circle are equal). Similarly, $$QD^2=QB^2$$. Since $OP$ and $OQ$ are both perpendicular to $PQ$ (the line joining the points of tangency), and $O$ is the center of the circumcircle of $ABCD$, it follows that $$OP=OQ$$This is because the perpendicular from the center of a circle to a chord bisects the chord, and here $PQ$ acts as a "chord" with respect to the "infinite radius" of the radical axis, implying $O$ is equidistant from $P$ and $Q$.

tapir1729 wrote: Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to $\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$. Show that $OP=OQ$. Merlijn Staps

tapir1729 wrote: Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to $\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$. Show that $OP=OQ$. Merlijn Staps $\angle APC=180-\angle CAP-\angle ACP=180-\angle ADE-\angle ACB-\angle CEB=180-\angle ADB-(180-\angle DBC)=\angle DBC-\angle ADB=\angle AFC$ so $F,A,C,P$ are concyclic. $\angle APF=\angle ACF=\angle ADB=\angle ADE=\angle CAP$ So $FP//AC$ but then if $O'$ is the center of $(FACP)$ we get that $OO'$ is the perpedicular bisector of $FP$ so $OP=OP=OQ$

We will show the power of $P,Q$ wrt $\omega = (ABCD)$ are equal. Let $X_A = PA\cap \omega, X_B = QB\cap\omega$. We call $\beta = \angle DAC, \alpha = \angle ADB$. We WLOG that $\beta > \alpha$. We have $\angle PAC = \angle QBD = \alpha$, $\angle CPA = \angle BQD = \beta - \alpha$. We get $\Delta CAP \sim \Delta DBQ$. We have $\frac{PA}{QB} = \frac{AC}{BD}$ so we want to show that $\frac{PX_A}{QX_B} = \frac{BD}{AC}$. Now since $\angle CX_AP = \angle ADC$ and $\angle DX_BQ = \angle DCB$. Note that this implies that $(CX_AP), (DX_BQ)$ have the same circumradius. So we have $\frac{\sin X_ACP}{\sin X_BDQ} = \frac{PX_A}{QX_B}$. Note that $\angle X_ACP = 180^\circ - \angle ADC - \beta + \alpha = \angle ACD + \alpha = \angle ABD + \alpha = 180 - \angle DAB = \angle DCB$. So we have that $\angle X_BDQ = \angle ADC$ which shows out desired result that $\frac{PX_A}{QX_B} = \frac{BD}{AC}$.

Extend $AD,BC$ and denote their intersection as $M$. Note $\angle(AP,AC)=\angle{APB}=\angle{ACB}; \angle{DAC}=\angle{DBC}=\angle{DCA}, \angle{PAM}=\angle{PCM}$, so $APMC$ is cyclic; since $\angle{PAC}=\angle{MCA}$, the shape is an isosceles trapezoid. Analogously, $BDMQ$ is also an isoscles trapezoid. Since $O$ lies on the perpendicular bisector of $AC, BC$, it also lies on the perpendicular bisector of $PM, QM$, so $OP=OQ=OM$

Let $AD \cap BC = F$. Then notice that $FPCA$ is cyclic as $\angle PAF = 180^\circ - \angle PAE - \angle EAB = \angle AEB$ and similarly $\angle PCF = \angle DEC$ as desired. Then we get $\angle APF = \angle ACF = \angle ACD = \angle ABD = \angle PAE$ so $AC \parallel PF \implies ACPF$ is an isosceles trapezoid. So then $AC$ and $FP$ share the same perpendicular bisector which implies that the perp. bisector of $FP$ passes through $O$. Analogously, we get that the perp. bisector of $FQ$ passes through $O$ so $OP = OQ$ as desired.

Let $AP$, $CP$, $BQ$, $DQ$ intersect $(ABCD)$ at $P'$, $P''$, $Q'$, $Q''$. Note that $\measuredangle Q'BD = \measuredangle BCA = \measuredangle BDA$, so $Q'A \parallel BD$. Similarly we get $P'B \parallel CA$, $P''D \parallel CA$ and $Q''C \parallel BD$. In particular, $\triangle PCA \sim \triangle QDB$. On the other hand, $\measuredangle QQ'D = \measuredangle BQ'D = \measuredangle P''CP' = \measuredangle PCP'$, so $\triangle PP'C \sim \triangle QDQ'$ as well. Since $P'C = AB = Q'D$, these triangles are actually congruent. Therefore, \[ PP' \cdot PA = QD \cdot PA = QB \cdot PC = QB \cdot QQ', \]so $P$ and $Q$ have the same power with respect to $(ABCD)$ and this gives $OP = OQ$.

Intersect $AP,BQ,CP,DQ$ with $ABCD$ again at $B',A',D',C'$. Claim: $OAB'CD'P$ is congruent to $OC'DA'BQ$ which would finish the problem. Observe that by angle chasing with directed angles that $\measuredangle B'AD=\measuredangle AED=\measuredangle ADC'$ so $C'DAB'$ is an isosceles trapezium. In particular, we get the direct congruence $\triangle OC'D\cong\triangle OAB'$. Similarly one gets $\triangle ODA'\cong\triangle OB'C$ and $\triangle OA'B\cong\triangle OCD'$. and piecing this together gives $OAB'CD'\cong OC'DA'B$. Finally, intersecting $AB'\cap CD'=P$ and $C'D\cap A'B=Q$ gives the desired congruence.

Let $PA$, $QB$, $PC$ and $QD$ meet $(ABCD)$ again at $S$, $T$, $U$ and $V$. It is easy to get by angle chasing that $\widehat{VB}$ = $\widehat{AU}$ = $\widehat{DC}$ and $\widehat{BA}$ = $\widehat{TD}$ = $\widehat{CS}$. (For example, $\widehat{VB} = 2\measuredangle VDB = 2\measuredangle DAE = \widehat{DC}$) Now, consider the rotation $\tau$ about $O$ such that $S \mapsto A$. By the equal arcs shown above we get $\tau(V) = A$, $\tau(B) = U$. $\tau(T) = C$ and $\tau(D) = S$. Finally \[ \tau(Q) = \tau(BT \cap DV) = UC \cap SA = P \]And hence, $P$ is just a rotation of $Q$ about $O$. $\blacksquare$

Nice and easy! Define $Q'$ as the intersection of tangents to $\omega_{1}$ at $A$ and $D$ and $P'$ as the intersection of tangents to $\omega_{2}$ at $B$ and $C$. Claim 1: $\color{blue}{P,P',Q',Q}$ are concylic. Proof: Let $\angle{DEA}=\angle{BEC}=\theta$. Then by alternate segment theorem, $\angle{Q'AD}=\angle{Q'DA}=\angle{DEA}=\theta$ and $\angle{P'BC}=\angle{P'CB}=\angle{BEC}=\theta$. Thus by exterior angle property, $\angle{QQ'A}=2\theta \implies \boxed{\angle{QQ'P}=2\theta}$ and $\angle{PP'B}=2\theta \implies \boxed{\angle{PP'Q}=2\theta}$. Thus, $\angle{PP'Q}=\angle{PQ'Q}$ implying $P,P',Q',Q$ are concylic as desired. $\blacksquare$ Claim 2: $\color{blue}{O \in (PP'Q'Q)}$ Proof: Note that we have $Q'A=Q'D$ implying $Q'$ lies on the perpendicular bisector of $AD$. But $O$ also lies on the perpendicular bisector of $AD$ thus, $OQ'$ is the perpendicular bisector of $AD$. Analogously, $OP'$ is the perpendicular bisector of $BC$. Thus, $\angle{AQ'O}=\angle{BP'O}=90^{\circ}-\theta$. Hence, $\angle{QQ'O}=2\theta+90^{\circ}-\theta=90^{\circ}+\theta$ and $\angle{QP'O}=90^{\circ}-\theta \implies \angle{QQ'O}+\angle{QP'O}=180^{\circ}$ implying $O \in (PP'Q'Q)$ as desired. $\blacksquare$ Claim 3: $\color{blue}{OP=OQ}$ Proof: We have, $\angle{OQP}=\angle{OQ'P}=\angle{OQ'A}=90^{\circ}-\theta$ and $\angle{OPQ}=\angle{OP'Q}=\angle{OP'B}=90^{\circ}-\theta \implies \angle{OQP}=\angle{OPQ} \implies \boxed{OP=OQ}$ as desired. $\blacksquare$

Actually, it is very easy exercise for projective transformations. There is a well known Theorem : if there are given circle and point $P$ in this circle, then there exist projective tranformtion that sends $P$ to $O$ where $O$ is the center of first circle, and sends first circle to the circle with same center. Solution: Just send $E$ to $O$ by theorem then we get configuration where $AC$ and $BD$ are diameters of this circle. Then points $P$ and $Q$ will be points at infinity, which gives us that $OP=OQ$.