Let ABCABC be a triangle with incenter II. The incircle ω\omega of ABCABC is tangent to the line BCBC at point DD. Denote by EE and FF the points satisfying AI∥BE∥CFAI \parallel BE \parallel CF and ∠BEI=∠CFI=90°\angle BEI = \angle CFI = 90°. Lines DEDE and DFDF intersect ω\omega again at points E′E' and F′F', respectively. Prove that E′F′⊥AIE'F' \perp AI.