$\angle ABC = \angle AKC = 90^\circ$, Find length of $BK$. 
Diagonal $AC$ of square $ABCD$ coincides with the hypotenuse of right triangle $ACK$. Assuming $B,K$ lie on the same side of $AC$, Prove the following - 
  
  
*
  
*$BK = \frac{\left| AK - CK \right|}{\sqrt{2}}$
  
*$DK = \frac{\left| AK + CK \right|}{\sqrt{2}}$     
  

It is easy to see that point $A,B,C,D,K$ lie on same circle with Diameter $AC$. This seems easy. But I am missing something. Any hint will be helpful. 
 A: If everything else fails, use coordinates. If you select a coordinate system where $A=(-1,0)$, $C=(1,0)$, $D=(0,-1)$ things are fairly manageable. We then have $K=(x,y)$ with $x^2+y^2$ and after a bit of algebra things work out as
$$ |DK|^2 = 2+2y $$
$$ \frac{(|AK|+|CK|)^2}{2} = 2+2\sqrt{1-x^2} $$
which are clearly the same.
A: Here is a synthetic proof:
Let $\omega$ be the circle circumscribed around the square $ABCD$. Since $$\angle \, AKC = \angle \, ABC = \angle \, CDA = 90^{\circ}$$ the point $K$ also lies on that circle. Let us assume that $C$ lies on the arc $BC$ not containing the points $A$ and $D$. Therefore the quads $ABKC$ and $AKCD$ are convex quads and both of them are inscribed in the circle $\omega$. Now, apply Ptolemy's theorem to $ABKC$ and obtain the identity
$$BK \cdot AC  + AB \cdot CK = AK \cdot BC$$ which by the fact that $BC = AB$ and $AC = AB \sqrt{2}$ becomes 
$$BK \cdot AB \sqrt{2}  + AB \cdot CK = AK \cdot AB$$ Canceling out $AB$ on both sides of the latter identity leads to
$$BK\sqrt{2}  + CK = AK$$ i.e.
$$BK = \frac{AK - CK}{\sqrt{2}}.$$ 
Next, apply Ptolemy's theorem to $AKCD$ and obtain the identity
$$CK \cdot DA + AK \cdot CD = DK \cdot AC$$ which by the fact that $DA = CD$ and $AC = CD \sqrt{2}$ becomes 
$$CK \cdot CD  + AK \cdot CD = DK \cdot CD \sqrt{2}$$ Canceling out $CD$ on both sides of the latter identity leads to
$$CK + AK = DK \, \sqrt{2}$$ i.e.
$$DK = \frac{AK+CK}{\sqrt{2}}.$$ 
The absolute values come from the alternative case when $K$ is on the arc $AB$ not containing the points $C$ and $D$.  
