In a square ABCD, F and E are any points on CD and CB respectively such that, AF = DE. Prove that angle EPF = 90 degrees.

Please please help me in this sum. My head is breaking.

given: ABCD is a square. and AF=DE

TPT: ∠EPF=90 deg.

proof: in ΔADF and ΔDCE

AD=CD and hypotenuse AF=DE

∠ADF=∠DCE=90 degree

therefore by the right angle hypotenuse side congruency 

ΔADF and ΔDCE are congruent triangles.

let ∠EDC=θ, by the congruent property ∠DAF will also be θ.

∠ADP=∠ADF-∠PDF=90-θ (since∠PDF is same as ∠EDC)

now in the triangle ADP:

∠APD=180-(∠DAP+∠ADP)=180-(θ+90-θ)=90

therefore ∠EPF=∠APD=90 degree(opposite angles are equal)