A contest geometry problem - a triangle and center of an arc

Question

Let $X$ be the center of an arc $ACB$ of circle circumscribed on the triangle $ABC$, where $|AC|<|BC|$. From the point $X$, drop line perpendicular to $BC$, and let the intersection of that line and $BC$ be $K$.

Prove that $|AC|+|CK|=|KB|$.

I must say, I don't have any clever idea for the time being. Any hints greatly appreciated.

Answer 1

Let $D$ be chosen on the line $BC$ so that $DC = AC$ and $C$ is inside the segment $BD$. Then triangle $ACD$ is isosceles. Since $X$ is the midpoint of the arc $ACB$, the line $CX$ is the exterior angle bisector of angle $\angle \, ACB$ and is thus the (interior) angle bisector of angle $\angle \, ACD$. As an angle bisector of the isosceles triangle $ACD$, the line $CX$ is also the orthogonal bisector of edge $AD$. Therefore, $XA = XD$. As mentioned before, $X$ is the midpoint of arc $ACB$ so $XA = XB$. Hence, $XA = XB = XD$ and thus triangle $BDX$ is isosceles. Then $XK$ is by assumption its altitude, so it is also the orthogonal bisector of $BD$. Hence, $DK = KB$. However, by construction, $DC = AC$ so $$KB = DK = DC + CK = AC + CK$$

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Now, just for fun, let us take a look at point $X'$, which is the diametrically opposite to point $X$. Then $X'$ is in fact the midpoint of arc $AB$, not containing point $C$. Let $K'$ be the orthogonal projection of $X'$ on $BC$. Analogously to the previous case, let $D'$ be chosen on the line $BC$ so that $AC = CD'$ and $D'$ is inside the segment $BC$. Then triangle $ACD'$ is isosceles. Since $X'$ is the midpoint of the arc $AB$, not containing $C$, the line $CX'$ is the angle bisector of angle $\angle \, ACB$. As an angle bisector of the isosceles triangle $ACD'$, the line $CX'$ is also the orthogonal bisector of edge $AD'$. Therefore, $X'A = X'D'$. As mentioned before, $X'$ is the midpoint of arc $AB$, so $X'A = X'B$. Hence, $X'A = X'B = X'D'$ and thus triangle $BD'X'$ is isosceles. Then $X'K'$ is its altitude, so it is also the orthogonal bisector of $BD'$. Hence, $AK' = K'D'$. However, by construction $AC = CD'$ so $$K'B = K'D' = CK' - CD' = CK' - AC$$

Answer 2

Let us assume that the circumradius of $ABC$ is $R$. In such a case we have $AC=b=2R\sin B$.
Since $X$ lies on the perpendicular bisector of $BC$, $$ AX = 2R\sin\widehat{XBA} = 2R\sin\left(\frac{\pi-A}{2}-B\right)$$ due to $\widehat{BXC}=A$. It follows that: $$ AK = AX\cos\left(\frac{\pi-A}{2}\right) $$ and $CA+AK=KB$ boils down to a simple trigonometric identity.