Hello!A solution for the first part now that the errors are corrected. So,we shall prove that $AB+AC=2BC\Rightarrow PQ\perp E_2F_2$. Suppose WLOG that $AB<AC$ Lemma:$OI\perp AI$.This one is proved here for example. Now,since $EF$ is the polar of $A$ with respect to the incircle we deduce that $EF\perp AI$. Using the lemma,this gives $EF\parallel OI$ which means that $OIFE$ is a trapezoid. Thus $PQ$ is its median whence we get $PQ\parallel EF\Leftrightarrow PQ\perp AI$. Hence,it suffices to show that $AI\parallel E_2F_2$. We will go on with some ugly computations in order to show that $AE_2=AF_2$. First of all $s=\frac{3BC}{2}$ (the semiperimeter). Obviously $AF_1=AE_1=2AF=2(s-BC)=2s-2BC=BC$.Thus $BF_1=BC-AB=AC-BC$. Also $CE_1=AC-BC$. We also have $BD=s-AC$ and $CD=s-AB$.The power of point gives $BF_1\cdot BF_2=BD^2\Leftrightarrow BF_2=\frac{(s-AC)^2}{AC-BC}=\frac{\frac{9BC^2}{4}-3AC\cdot BC+AC^2}{AC-BC}$ Thus $AF_2=AB+BF_2=\frac{\frac{9BC^2}{4}-3AC\cdot BC+AC^2+AB\cdot AC-AB\cdot BC}{AC-BC}$. Also $CD^2=CE_1\cdot CE_2\Leftrightarrow CE_2=\frac{(s-AB)^2}{AC-BC}\Leftrightarrow$ $AE_2=\frac{(s-AB)^2}{AC-BC}-AC=\frac{\frac{9BC^2}{4}-3AB\cdot BC+AB^2-AC^2+AC\cdot BC}{AC-BC}$ Now,it suffices to show that $\frac{9BC^2}{4}-3AB\cdot BC+AB^2-AC^2+AC\cdot BC=\frac{9BC^2}{4}-3AC\cdot BC+AC^2+AB\cdot AC-AB\cdot BC$. This is equivalent to $4AC\cdot BC-2AB\cdot BC+AB^2-AB\cdot AC-2AC^2=0\Leftrightarrow$ $2BC(2AC-AB)+(AB+AC)(AB-2AC)=0\Leftrightarrow (2BC-AB-AC)(2AC-AB)=0$ which is true. Thus the triangle $AE_2F_2$ is isosceles with $AE_2=AF_2$ and $AI$ is the external bisector of $\angle{A}$ whence we easily get $AI\parallel E_2F_2$ which is what we wanted to prove. $\rule{430pt}{1pt}$ I haven't put $P,Q$ in the diagram because they aren't neededn for the proof. The converse seems a lot more difficult.If I solve it before someone else,I'll upload my solution.

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My solution: Lemma : Let $ A, B, C, D $ (in order) be four points on a line s.t $ (A,C;B,D)=-1 $ . Let $ B^* $ be the reflection of $ B $ in $ C $ and $ X $ be the midpoint of $ AC $ . Then $ CD^2=B^*D \cdot XD $ . Proof : From $ (A, C; B, D)=-1 \Longrightarrow AB \cdot CD=AD \cdot BC \Longrightarrow 2 \cdot CD \cdot BC+AC \cdot BC=AC \cdot CD $ , so $ 2 \cdot CD^2=(CD-BC) \cdot (2 \cdot CD+AC)=2 \cdot B^*D \cdot XD \Longrightarrow CD^2=B^*D \cdot XD $ . ____________________________________________________________ Back to the main problem : Let $ E_3=DF \cap AC, F_3=DE \cap AB $ . Since $ (E_3,E;A,C)=-1 $ , so from $ CE^2=CD ^2=CE_1 \cdot CE_2 $ and the lemma $ \Longrightarrow E_2 $ is the midpoint of $ EE_3 $ , hence we get $ E_2E^2=E_2A \cdot E_2C \Longrightarrow E_2 $ lie on the radical axis of $ \{ \odot (I), \odot (O) \} $ . Similarly we can prove $ F_2 $ lie on the radical axis of $ \{ \odot (I), \odot (O) \} \Longrightarrow E_2F_2 $ is the radical axis of $ \{ \odot (I), \odot (O) \} $ . ... $ (\star) $ From $ (\star) $ we get $ E_2F_2 \perp IO $ , so $ PQ \perp E_2F_2 \Longleftrightarrow IO \parallel PQ \Longleftrightarrow IO \perp AI \Longleftrightarrow AB+AC=2BC $ . (for the proof of the last step you can see the link gavrilos gave above or here ) Q.E.D

The configuration is very easy for calculation. Notice the well-known lemma $AB + AC= 2BC \iff AI \perp OI$ (this can be proved by Ptolemy) Also note that $OI \parallel FE \iff OI \parallel PQ$. This motivates us to check if $E_2F_2 \perp IO$. Brute force calculations show that $E_2E^2 = E_2A \cdot E_2C$ and $F_2F^2 = F_2B \cdot F_2A$. All calculations can be done with lengths $AB, BC, CA$ only. This shows that $E_2$ and $F_2$ lie on the radical axis of the incircle and circumcircle, so $E_2F_2 \perp IO$ always. Therefore, $E_2F_2 \perp PQ \iff OI \parallel PQ \iff OI \parallel EF \iff AI \perp OI \iff AB+AC=2BC$.