<Qua-2025-751>, hint

Let B' on line DC, such that CB'=AB

AD^2-CD^2=ABxCD---> AD^2=(AB+CD) CD=(AB+CD) FD  which means triangle B'AF is right triangle , <FAB'=90. Triangles EDC and EAB are similar

DC/CB'=DE/EA so that CE//B'A , <B'AF=90 Hence CE is perpendicular to AF