Here is a solution based only on angle chasing. First, note that $CL$ is $C-\text{symmedian}$ in $\triangle ACB$ and $DL$ is $D-\text{symmedian}$ in $\triangle ADB$. We have that $\angle DEK=\angle DEC= \angle DAC= \angle DBC$. Now we hunt after $\angle DKE$, if we show that $\angle DKE=180^{\circ}-2 \angle DAC$ then we'd be done. Now using the given parallel lines and since $DCFL$ is cyclic we get \begin{align*} \angle DKE &= 180^{\circ} -\angle FDM - \angle DFC \\ &= 180^{\circ}- \angle DMC-\angle DFC \\ &=180^{\circ}-\angle DMC- \angle DLC. \end{align*}Chasing $DLC$ using the symmedian properties we have \begin{align*} \angle DLC &= \angle DMC - \angle MDL - \angle MCL \\ & = \angle DMC-(\angle ADB-2 \angle BDM)-(\angle ADB-2\angle ACM) \\ &=\angle DMC- 2\angle ADB+2 \angle BDM +2\angle ACM. \end{align*} Now going after $\angle DMC$ we get \begin{align*} \angle DMC &= 180^{\circ}-\angle CDM -\angle DCM \\ &= 180^{\circ} - \angle D +\angle ADM - \angle C+ \angle BCM \end{align*}Finally we go after $\angle DMC+\angle DLC$ using that $ABCD$ is cyclic to get \begin{align*} \angle DMC+\angle DLC &=2 (\angle DMC- \angle ADB +\angle BDM +\angle ACM) \\ &= 2(180^{\circ} - \angle D +\angle ADM - \angle C+ \angle BCM - \angle ADB +\angle BDM +\angle ACM) \\ &=2(180^{\circ}- \angle D- \angle C+\angle ADB) \\ &=2 \angle DBC \end{align*} So we have $\angle DKF=180^{\circ}-\angle DMC- \angle DLC=180^{\circ}-2 \angle DBC=180^\circ-2 \angle DEK$ wich implies that $ \triangle DKE$ is isosceles with $DK=KE$. $\blacksquare$

DM,DL-(O)=G,H so GH//BC, FDE=HCL=MCG so CG//DE so KE=KD

nguyenhaan2209 wrote: DM,DL-(O)=G,H so GH//BC, FDE=HCL=MCG so CG//DE so KE=KD Your answer is perfect！ But there is a little problem.GH//BA

nguyenhaan2209 wrote: DM,DL-(O)=G,H so GH//AB, FDE=HCL=MCG so CG//DE so KE=KD Here is a more detailed explanation of the above solution: $\angle FDE=$ $=\angle FDL-\angle EDL=$ $=\angle FCL-\angle EDH=$ $=\angle ECL-\angle ECH=$ $=\angle HCL=$ $=\angle ACL-\angle ADH=$ $=\angle BCM-\angle BCG=$ $=\angle MCG$ $\therefore DE\parallel CG$ $\therefore \overarc{DC}=\overarc{EG}$ $\therefore \angle DEC=\angle EDG$ $\therefore \angle DEK=\angle EDK$