Let

be the circumcircle of a triangle

. Denote by

and

the midpoints of the sides

and

, respectively, and denote by

the midpoint of the arc

of

not containing

. The circumcircles of the triangles

and

intersect the perpendicular bisectors of

and

at points

and

, respectively; assume that

and

lie inside the triangle

. The lines

and

intersect at

. Prove that

.
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Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA = KT$.