Let $B'$ be the antipode of $B$ WRT the circle $(O)$ and define $X' \equiv PB' \cap AC.$ By Pascal theorem for degenerate cyclic hexagon $PPB'BAC,$ it follows that intersections $E \equiv PP \cap AB,$ $X' \equiv PB' \cap AC$ and $O \equiv BB' \cap PC$ are collinear $\Longrightarrow$ $X$ and $X'$ coincide. Therefore, $XPB' \parallel BCY$ $\Longrightarrow$ $O$ is the midpoint of $XY.$ With the same argument we'll have that $O$ is the midpoint of $UV$ $\Longrightarrow$ $XVYU$ is a parallelogram $\Longrightarrow$ $XV=YU.$

Quote: A quadrilateral $ABCD$ is inscribed into a circle with center $O.$ Points $P$ and $Q$ are opposite to $C$ and $D$ respectively. Two tangents drawn to that circle at these points meet the line $AB$ in points $E$ and $F.$ ($A$ is between $E$ and $B$, $B$ is between $A$ and $F$). The line $EO$ meets $AC$ and $BC$ in points $X$ and $Y$ respectively, and the line $FO$ meets $AD$ and $BD$ in points $U$ and $V$ respectively. Prove that $XV=YU.$ $\textbf {Lemma}:$ Let $\triangle {ABC}$ inscribed circumcenter $(O)$ and $P_a$ be the symmentric point of $A$ with respect to $O$.The tangent at $P_a$ intersects $BC$ at $K$. Let $KO$ intersects $AB$ and $AC$ respectively at $M$ and $N$. Prove that $O$ is the midpoint of $MN$. Proof. Let the second tangent $Kx$ from $K$ to $(O)$ intersects $(O)$ at $P'_a$. Then $\angle {xP'_aA}=$ ${\angle P'_aP_aA}$ $=\angle P'_aKO$, which implies $P'A$ $\|$ $KO$. However, since $P'_aBP_aC$ is a harmonic quadrilateral, thus $(AB, AC, AP_1, AP'_a)=-1$. Or in other words, $(AM, AN, AO, AP'_a)=-1$. As a result, $O$ is the midpoint of $MN$. Back to our problem, by the lemma, we have known that $O$ respectively is the midpoint of $XY$ and $UV$. Thus, $XVYU$ is a parallelogram. Therefore, $XV=YU$. Our proof is completed then. $\square$

mathVNpro wrote: $\mbox{\textbf {Lemma:}}$ Let $\triangle {ABC}$ inscribed circumcenter $(O)$ and $P_a$ be the symmentric point of $A$ with respect to $O$.The tangent at $P_a$ intersects $BC$ at $K$. Let $KO$ intersects $AB$ and $AC$ respectively at $M$ and $N$. Prove that $O$ is the midpoint of $MN$. I proved exactly the same thing - but in a different way: Let $l$ be the line perpendicular to $KO$ through $K$ and denote the intersections of $l$ with $P_aB$ and $P_aC$ by $X$ and $Y$, respectively. Then, by the butterfly theorem, we know that $XK=KY$. As $\sphericalangle{YKN}=\sphericalangle{YCN}=\frac{\pi}{2}$ and $\sphericalangle{XKM}=\sphericalangle{XBM}=\frac{\pi}{2}$, $YKCN$ and $XKMB$ are cyclic, implying $\sphericalangle{P_aXK}=\sphericalangle{BXK}=\sphericalangle{KMA}=\sphericalangle{NMA}$ and $\sphericalangle{P_aYK}=\sphericalangle{CYK}=\sphericalangle{CNK}=\sphericalangle{ANM}$, i.e. the triangles $\triangle{P_aXY}$ and $\triangle{AMN}$ are similar. As furthermore $\sphericalangle{KP_aY}=\sphericalangle{P_aAC}=\sphericalangle{OAN}$, we know that $\frac{MO}{ON}=\frac{XK}{KY}=1$ or $MO=ON$ as required.