Iran TST 2014 third exam - Problem 6

Problem: The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the side $BC$ at $D$. Let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$.
Prove that: $\angle{BAD}=\angle{CAX}$.
Solution: 
Let $\angle{CBA}=2a,\angle{BCA}=2b$.
Let $Q,W,R$ are foot of perpendicular from $E,M,F$ on $BC$.
$MB=MC\implies BW=WC$, but $M$ is midpoint of $EF\implies WQ=WR\implies BQ=CR\implies BE.cos(a)=CF.cos(b)$.
Let $I_a$ is $A-excenter$.
Let $T,S$ are foot of perpendicular from $X$ to $BI_a,CI_a\implies \frac{XT}{XS}=\frac{BE}{CF}=\frac{cos(b)}{cos(a)}=\frac{sin\angle{BCI_a}}{sin\angle{CBI_a}}=\frac{sin\angle{XI_aB}}{sin\angle{XI_aC}}\implies XI_a$ is symmedian of triangle $CI_aB$.
Let $L$ is midpoint of arc $BAC$, $\angle{LBC}=\angle{LCB}=\angle{CI_aB}\implies LB,LC$ are tangent to circumcircle of triangle $CI_aB$.
$\frac{BX}{CX}=\frac{sin\angle{BLX}}{sin\angle{CLX}}=(\frac{cos(b)}{cos(a)})^2(1)$.
$\frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{AC.BD}{AB.CD}=\frac{tan(b)}{tan(a)}.\frac{sin(2a)}{sin(2b)}=(\frac{cos(a)}{cos(b)})^2(2)$.
$(1)(2)\implies \frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{sin\angle{CLX}}{sin\angle{XLB}},(\angle{BAD}+\angle{CAD}=\angle{CLX}+\angle{BLX})\implies \angle{CAX}=\angle{CLX}=\angle{BAD}$.
So we are done.
Source: https://artofproblemsolving.com/community/c6h590555p3497371