Redefine $P$ the midpoint of arc $BAC$ and $YP \cap AC = \{Z\} $ Hence arcs $BP$ and $CP$ are equal so $\angle CYZ = \angle CBP = \angle BUP$ Notice that $YB \parallel AC $ so $\angle BYZ = \angle YZC = \angle CYZ $ so $CY = CZ = CX$ so $X \equiv Z $ and we are done.

parmenides51 wrote: Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$, such that $CX = CY = AB$. (The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $\omega$ for the second time in point $P$. Show that $PB = PC$. by Iman Maghsoudi $$\angle PBC=\angle XYC=\angle YXC=(\smile AP+\smile CY)/2=(\smile AP+\smile AB)/2=\smile BP/2=\angle BCP.$$As I know, it is the problem from Iranian geometric olimpiad.

Let $Q$ be another point on $\omega$ so that $CQ=AB$. With $CX=CY=CQ$ by Fact 5 $X$ - incenter of $AYQ$, so $AP=PQ$. $P$ - midpoint of the arc $BAC$.

$\angle XYC=\angle YXC=\alpha$ $\angle ABP=\angle ACP=\beta$ $\angle PBC=\alpha$ if $\angle BCA=\alpha-\beta$ we are done note that $ABYC$ is isosceles trepazoid $\angle XCY= 180-2\alpha$ $\implies$ $\angle YBC=\alpha-\beta$ and $\triangle BYC $ has two known angle and we can use them for find the third so, $\angle BCY=180-3\alpha-\beta$ $\implies$ $\angle BCA=\alpha-\beta$ as we desired

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