My teacher said that it is a problem from Russian 2011. Let $E$ is the intersection of $(I)$ with $AC$. Easy to see that quadrilateral $FHED$ and $FEKD$ is harmonic.Use harmonic quadrilateral propeties and Ptoleme theorem we have So we have $2EK.FD=2EF.DK=ED.FK$ and $2HF.DE=2HE.DF=HD.EF$ multiply two side of each indentity : $4.(HF.DE).(EF.DK)=(HD.EF).(ED.FK)$ or $HD.FK=4FH.DK$ Use again Ptoleme theorem for $FHDK$ we imply $FD.HK=3 FH.DK$. \GED/.

Could you explian it more detail? because i don't know how harmonic is work?

A harmonic quad $ABCD$ inscribed in a circle satisfies that $AB \cdot CD = AD \cdot BC =\frac{1}{2} \cdot AC \cdot BD$, such quad has a characteristic : the tangent lines through A,C and line BD concurrent, and wiseversa, this can be easily proved. For this problem, It's obvious that $HK$ is not calculatable(or it'll too difficult), so just apply Ptolemy's theorem to change the proposition to $\frac{HF \cdot DK}{HD \cdot FK}=\frac{1}{4}$, and then apply the conclusion of harmonic quad, as Love_Math1994 stated.