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.
On
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.
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 $D$.