In triangle CDE, point K lies on the side CE, and the angle CKD is acute. Prove that DE> DK

In the condition of the problem, a triangle CDE is given. Connect the point K, which lies on the CE side, with the angle CDE. Consider a triangle KDE, the angle DKE is adjacent to the angle CKD, which is acute, so the angle DKE is obtuse. A triangle can have only one obtuse angle. The angle of the triangle KDE DEK and the angle KDE are sharp. An obtuse angle is larger than an acute one. Against the larger angle lies the larger side. The DE side lies opposite the larger DKE angle, and the DK side lies opposite the smaller DEK angle. So the DE side is larger than the DK side.
Answer: DE> DK.