Prove that $EF$ perpendicular to $OI_a$

Let $O, I$ and $I_a,$ denote the circumcenter,incenter and excenter in the angle $A$ of a triangle $ABC$. $BI$ meets $AC$ at $E$. $CI$ meets $AB$ at $F$. Prove that $EF$ perpendicular to $OI_a$

It is somewhat like this problem:
Let $H$ be the orthocenter of a triangle $ABC$, and $X, Y, Z$ be the feet of the altitudes from $A, B, C$. $XZ$ meets $HB$ at $E$. $XY$ meets $HC$ at $F$. $O_E$ is center of euler circle of triangle $ABC$. Prove that $AO_E$ perpendicular to $EF$

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The full answer is on the picture.

the solution is calculate the angle related EF, since EF is the foot of bisector, we can find all the ratios.

the most difficult part is to calculated the angle of $I_{a}O$, you have to know all staff about excircles. for tangent point, I had proof it in SE ,see here.

you have to know $R=\dfrac{r_{a}+r_{b}+r_{c}-r}{4}$,it has same difficult as this problem if you haven't proof it before.

another is that $r_{a}*(s-a)= r_{b}*(s-b)=r_{c}*(s-c)=rs=\sqrt{s(s-a)(s-b)(s-c)}$,here $s=\dfrac{a+b+c}{2}$.

for $QM=\dfrac{r_{a}-r}{2}$, I didn't proof it as it is not difficult.hint:M is middle point.

The rest work is heavy calculations.

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the polar axis of $\triangle II_bI_c$ is $EF$, $OI_a$ is the Eulur line.

Recall that the polar axis of a triangle perpendicular to the Eulur line, Clearly, $EF\perp OI_a$

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