Let $\angle BAI=\alpha$ and $\angle ABI=\beta$. After some experiments it is easy to see that $\angle KIF=2\alpha$ and $\angle KIE=2\beta$. Therefore, $\frac{KF}{KE}=\frac{\frac{KF}{IK}}{\frac{KE}{IK}}=\frac{\frac{sin2\alpha}{sin\angle IFK}}{\frac{sin2\beta}{sin\angle IEK}}=\frac{sin2\alpha}{sin2\beta}$ Let $\angle IMD=x$, then $\angle IDM=90^{\circ}-\alpha-x$. $\frac{DI}{IM}=\frac{sinx}{sin(90^{\circ}-\alpha-x)}=\frac{EI}{IM}=\frac{sin\beta}{sin(90^{\circ}-\alpha-\beta)}$ Therefore, $\beta=x$ because $\beta>x$ or $\beta<x$ is a contradiction. Similarly we have $\angle DNI=\alpha$ which implies that $\frac{DM}{DN}=\frac{sin2\alpha}{sin2\beta}=\frac{KF}{KE}$. Therefore, $KE.DM=KF.DN$

Since points $D$ and $E$ are symmetrical wrt line $BI$ and points $F$ and $D$ are symmetrical wrt line $AI$ then $DN=NE$ and $DM=MF$. Let $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{ACB}=\gamma$. Then $\angle{EBN}=\frac{\beta}{2}$, $\angle{BEN}=180^{\circ}-\angle{CEF}=90^{\circ}+\frac{\gamma}{2}$, $\angle{INF}=\angle{BNE}=180^{\circ}-(90^{\circ}+\frac{\beta+\gamma}{2})=180^{\circ}-(90^{\circ}+\frac{180^{\circ}-\alpha}{2})=\frac{\alpha}{2}=$ $=\angle{IAF}.$ Thus point $N$ lies on the circumcircle of $ADIF$. By analogiousy, point $M$ lies on the circumcircle of $BDIE$. Since line $EF$ intersects these circles, we have \[ KN \cdot KF=KD \cdot KI=KM \cdot KE \implies KF \cdot (NE - KE)=KE \cdot (MF-KF) \implies \implies KF \cdot NE=KE \cdot MF \implies KF \cdot DN=KE \cdot DM. \] as desired.

As shown above, $M$ belongs to circle $\odot (BEID)$, $N$ belongs to circle $\odot (FIEC)$, hence $\triangle KDM\sim\triangle KEI$ and $\triangle KDN\sim\triangle KFI$, so we get $\frac{DM}{IE}=\frac{DK}{EK}$ and $\frac{DN}{IF}=\frac{DK}{FK}$; dividing the two relations side by side we get $\frac{DM}{DN}=\frac{FK}{KE}$ (with $IE=IF$), hence done. Best regards, sunken rock

Dear Mathlinkers, you can also see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=491117 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=502163 Sincerely Jean-Louis

Lemma 1:$AM \perp BM$ and $AN \perp BN$. Proof:See Lemma 1 of my post at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=469024 Lemma 2: $\frac{FK}{EK}=\frac{BC}{AC}$. Proof:See the Lemma of my post at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=80872&&start=20 (You can avoid trig if you really want) Now note that $\angle{IDA}=\angle{INA}=90^{\circ} \implies ADIN$ concyclic.Thus $\angle{DNI}=\angle{DAI}=\frac{A}{2}$.Also note that $\angle{ANB}=\angle{AMB}=90^{\circ} \implies ANEB$ concyclic.So $\angle{BNM}=\angle{BAM}=\frac{A}{2}$.Combining these we get that $\angle{DNM}=A$.Similarly we will get $\angle{DMN}=B$.Thus $\frac{DM}{DN}=\frac{sinA}{sinB}=\frac{BC}{AC}$ and combining this with lemma 2 gives the desired result.

we first prove a lemma lemma = let $ABC$ be a triangle with incircle touching $AB,AC$ at $D,E$. let $I$ be incentre and let $CI$ intersect $DE$ at $G$. than $\angle CGB = 90$. proof = since $AD=AE$ we get $\angle AED = 90-A/2$ and hence $\angle CEG = 90+A/2$ which gives $\angle CGE =\angle IGE= B/2$ thus $BIGD$ is cyclic quad. and hence , $\angle CGB=\angle IGB = \angle IDG = 90$ thus proving the lemma. consequently we can prove $AMNB$ as a cyclic quad too. main proof = from above lemma we can immediately see that $\angle AMI = \angle IDA = 90$ and hence $ADIM $ and similarly $BDIN$ are cyclic quad. thus by sine law in triangle $IDM,IDN$ we get $\frac{DM}{sin BAM} = \frac{ID}{sin A/2}$ and $\frac{DN}{sin NBA}=\frac{ID}{sin B/2}$ thus, $\frac{DM}{DN} = \frac{sin BAM.sin B/2}{sin NBA.sin A/2}=\frac{cosB/2.sinB/2}{sinA/2.cosA/2}=\frac{sinB}{sinA}$ since $\angle NBA = 90-A/2$ and $\angle BAM = 90-B/2$ similarly applying sine law in triangles $FIK,EIK$ yield $\frac{KF}{KI} = \frac {sin FIK}{sin IFK}$ and $\frac{KE}{KI}=\frac{sin EIK}{sin IEK}$ thus $\frac{KE}{KF} =\frac{sin EIK}{sin FIK} = \frac{sin B}{sin A}$ since $\angle IEK =\angle IFK = C/2$ and $\angle FIK = A$ and $\angle EIK = B$ since $ADIF,DIEB$ are cyclic quad. thus we get $\frac{KE}{KF} =\frac{DN}{DM}$ or $DM.KE=DN.KF$ hence proved