Since $\triangle{IDE'}, \triangle{IDF'}$ are isosceles, $\angle{FIF'} = \angle{FID} - (180^{\circ} - 2\angle{IDF})$ and $\angle{EIE'} = \angle{EID}-(180^{\circ} - 2\angle{IDE})$, so it suffices to show that $\angle{FID} - \angle{EID} = 2(\angle{IDE} - \angle{IDF})$. If $M = AI \cap (ABC)$ then $ID, IM$ are isogonal wrt $\angle{BIC}$, so $\angle{FID} - \angle{EID} = 2\angle{DIM}$ and $2(\angle{IDE} - \angle{IDF}) = 2(\angle{BIM} - \angle{CIM})$, so they are clearly equal. $\square$

Consider that $\angle EBI=\angle ICD=\angle E'DB=\angle E'F'D=\angle IFD$ Hence $E'F'//EF$ Which means that $E'F'\bot AI$

No need to introduce any new points: Since $\overline{EIF} \perp AI$, it suffices to show that $EF \parallel E'F'$. Since $\angle E'F'D = \angle E'DB = \angle EDB$, this is equivalent to $\angle EDB = \angle EFD$. To prove this, notice that $BDIE$, $CFID$ are cyclic by Thales. Simple angle chasing hence gives $$\angle EDB = \angle EIB = \angle AIB - 90^\circ = \frac{1}{2} \angle ACB = \angle ICD =\angle IFD = \angle EFD.$$

Claim1: E , I and F are collinear .Proof : BE || AI || CF . $ \angle CFI $ = $ 90^\circ $ = $ \angle FIA $ and $ \angle BEI $ = $ 90^\circ $ = $ \angle EIA $ hence $ \angle EIA $ + $ \angle AIF $ = $ 180^\circ $ . hence E , I and F are collinear . As a result of the above proof : $ AI \perp EF $ .Now if we prove EF || E'F' , the problem is solved . It's enough to prove $ \angle FED = \angle F'E'D $ .Due to the symmetry of the problem, the result is obtained in a similar way $ \angle EFD = \angle E'F'D $ . Claim2: IEBD cyclic .Proof : $ \angle IDB $ = $ 90^\circ $ . (It is connected from the center of the circle to the tangent) Now $ \angle IDB $ + $ \angle IEB $ = $ 180^\circ $ . As a result, the desired sentence was proved. It results in a similar way IFCD cyclic . $ \angle IED = \angle IBD $ and $ \angle F'E'D = \angle CDF' = \angle FIC $ .hence it's enough to prove $ \angle IBD = \angle FIC $ . We know that $ \angle IBD $ + $ \angle ICD $ + $ \angle IAC $ = $ 90^\circ $ . (*)On the other hand $ \angle FIC $ + ($ 90^\circ $ - $ \angle ICD$) + $ 2 \angle ICD $ + ($ \angle FCA $ = $ \angle IAC $) =$ 180^\circ $ . hence $ \angle FIC $ +$ \angle ICD $ + $ \angle IAC $ = $ 90^\circ $ .Using (*) and (**), the proof is complete.

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