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1. Let $ω$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $ω$, such that $CX = CY = AB$.(The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $ω$ for the second time in point $P$. Show that $PB = PC$.

$\angle CAP = \angle CBP= \angle CYP= \angle CXY= \angle x$
$\angle BAC =\angle BPC =\angle a = \angle YCA$ [Cause, $YCAB$ is a cyclic quad, and $BA=YC$]
$\angle APB=\angle ACB=\angle b$
$\angle PCA=\angle PBA=\angle n$ [All of them by angle chasing]

Now, $180- (\angle a + \angle x)= \angle n+ \angle b$ [from $ \triangle APB$]

Again, $180- (\angle a + \angle x) = \angle x$ [from $ \triangle CXY$]

That means, $\angle n+ \angle b=\angle x$
So,$ \angle PCB=\angle PBC $
Or, $PB=PC$

Let ,
$\angle CPY = \angle CBY = \angle p$
$\angle PBC = \angle PYC = \angle q$

$CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$

$AB = CY , BY = BY , \angle BAY = \angle BCY$
So,$\triangle ABY \cong \triangle BCY$

Thus, $\angle CBY = \angle AYB = \angle ACB = \angle p$

From $\triangle PXC$, $\angle p + \angle PCX = 180 - \angle PXC$
But, $\angle YXC = 180 - \angle PXC = \angle q$.

$\angle p + \angle PCX = \angle q \Rightarrow \angle PCB = \angle PBC$

$\therefore PB = PC$

Here is the official solution :
We know that $CX = CY$ therefore:
$\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$ [But how?]
Also we have $AB = CY$ therefore $\widehat{AP} + \widehat{CY} = \widehat{AP} + \widehat{AB} = \widehat{PB}$
So $PB = PC$.

How we can write $\angle YXC = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$?

As, $ \angle YXC = \angle CPY + \angle PCA$. So ,$ \angle YXC = \angle XYC \Rightarrow \angle CPY + \angle PCA = \angle XYC \Rightarrow \widehat{AP} + \widehat{CY} = \widehat{PC}$