MathNet 2015-part-04 — 191 problems (EN–VI) Question 12

Let $H$ be the intersection point of the altitudes $AD$ and $BE$ of an acute triangle $ABC$. The circumcircle of the triangle $ABC$ intersects the circle with diameter $CH$ at the point $K$ other than $C$. Prove that
$$
\frac{DK}{KE} = \frac{DH}{HE}.
$$