<Tri-2024-2164>, hint

Let <A=90, inradii of triangles ABD and ACD be r and r' respectively

In triangle  <EDF=90, DE/DF=r/r'=AB/AC and so triangles BAC and EDF are similar

Hence <ABC=<DEF=<DAC ,so E, D, L, A are concyclic ---> <ALE=<ALK=<ADE=45

Since <A=90 <AKL=90-45=45.  Triangle AKL is isosceles. AK=AL.