<Qua-2025-801>, hint

 Let AK and DC meet at P.

Then <AKB=<CKP=<DAK=<A/2 (BC//AD)

<MPC=<MAL=<A/2 (AB//DP).  Thus CK=CP and triangles KAL and KPC are congruent

Since ALMN are cyclic <MNL=<MLN=<A/2 . 

Finally ML=MN=MC and <LNC=90.