<Cir-2024-776>, hint

Since <AKB=<IDB=90. BKDIF are concyclic. <IKD=<IBD =<B/2 

Hence <DKH=90-B/2. Similarly <ELH=90-A/2.

Suppose circles DKH and ELH meet at P, then <DPE=180-(A+B)/2

Since <DFE=(A+B)/2 FDPE are cyclic and P lies on incircle of ABC.