Let incircle touch $BC$ at $D$ and $\ell$ be polar of $X$ wrt incircle. It's well-known

that $DF, CX,$ and $\ell$ are concurrent (at $T$), similarly $DE, BX,$ and $\ell$ (at $S$). Considering all polars wrt incircle, we get $T$ lies on the polar of both $X$ and $B$. From La Hire get $BX$ is polar of $T$, which means $TI\perp BX$. Similarly $SI\perp CX \implies$ $I$ is orthocenter of $\triangle TSX$. As $X$ is always in $\triangle TSI$, we get $\angle BXC=\angle TXS>90^{\circ}$.

Let the incircle be the unit circle on the complex plane. Let $a$ be $E$, $b$ be $F$, and $c$ be the last contact point in $\triangle ABC$. Let $M$ be the midpoint of $BC$. We find $X$ is $\frac{a+b}{2}$ and $M$ is $$\frac{bc}{b+c}+\frac{ac}{a+c}.$$We want to prove $MX < MB$, which eventually simplifies to $$\frac{(a+b)^2(a-c)(b-c)}{4ab(a+c)(b+c)} > 0.$$ Anyone can continue?

Let $\angle BAC= \alpha$ $BC=a$ $AC=b$ $AB=c$ $\angle BXC = \varphi$ and $x=\frac{b+c-a}{2}$ it is well know that $AF=AE=x$ And also we get $XF=XE=x\sin(\frac{\alpha}{2})$ By law of cosine on triangles $AXC$ $AXB$ we get $$XC^2=x^2\sin^2\left(\frac{\alpha}{2}\right)+(b-x)^2-2x(b-x)\sin\left(\frac{\alpha}{2}\right)\cos\left(90+\frac{\alpha}{2}\right)$$ $$XB^2=x^2\sin^2\left(\frac{\alpha}{2}\right)+(c-x)^2-2x(c-x)\sin\left(\frac{\alpha}{2}\right)\cos\left(90+\frac{\alpha}{2}\right)$$ Hence $$XC^2=x^2\sin^2\left(\frac{\alpha}{2}\right)+(b-x)^2+2x(b-x)\sin^2\left(\frac{\alpha}{2}\right)$$ $$XB^2=x^2\sin^2\left(\frac{\alpha}{2}\right)+(c-x)^2+2x(c-x)\sin^2\left(\frac{\alpha}{2}\right)$$ And by law of cosine on $\triangle BXC$ $$a^2=XB^2+XC^2-2 \times XB \times XC \times \cos(\varphi)$$ Hence $$2 \times XB \times XC \times \cos(\varphi)=XB^2+XC^2-a^2$$ $$ 90<\varphi <180 \iff \cos(\varphi)<0 \iff XB^2+XC^2-a^2<0$$ Using law of cosine in $\triangle ABC$ $$a^2=b^2+c^2-2bc\cos(\alpha)$$ After simplifying we get $$XB^2+XC^2-a^2=2x^2\cos^2\left(\frac{\alpha}{2}\right)-2x(b+c)\cos^2\left(\frac{\alpha}{2}\right)+2bc\cos(\alpha)$$ Using the identity $\cos^2\left(\frac{y}{2}\right)=\frac{\cos(y)+1}{2}$ we get $$XB^2+XC^2-a^2=(cos(\alpha)+1)\left[x^2-x(b+c)+\frac{2bc\cos(\alpha)}{\cos(\alpha)+1}\right]$$ and since $\cos(\alpha)>-1$ it is enough to show that $$x^2-x(b+c)+\frac{2bc\cos(\alpha)}{\cos(\alpha)-1}<0 \iff \frac{(b+c-a)^2}{4}<\frac{(b+c)(b+c-a)}{2}+\frac{2bc(b^2+c^2-a^2)}{a^2-(b+c)^2} \iff ((b+c)^2-a^2-4bc)^2>0$$Note that the equality is when $\alpha =\pi $ or $\lvert b-c\rvert =a$ which is impossible since $ABC$ is non degenerate triangle Q.E.D

Here's a fun sol using pure trig bash Define \(a, b, c\) as the lengths of the sides \(BC\), \(CA\), and \(AB\), respectively, and \(\alpha, \beta, \gamma\) as the angles \(\angle CAB\), \(\angle ABC\), and \(\angle BCA\), respectively. Note that to prove \(\angle BXC > 90^\circ\), it is sufficient to prove that \(a^2 > BX^2 + CX^2\). We will use trig bash to determine the lengths of \(BX\) and \(CX\). Observe that \(AF = AE\), which implies that \(\angle AXF = 90^\circ\), and since \(IF = IE\), it follows that \(\angle IXF = 90^\circ\). Therefore, we know that \(A, X, F\) are collinear, implying that \(\angle FAX = \frac{\alpha}{2}\). Hence, we have \[ FX = AF \sin \frac{\alpha}{2} = (s-a) \sin \frac{\alpha}{2}. \] Also note that \[ \cos \angle BFX = -\cos \angle AFX = -\sin \frac{\alpha}{2}. \] By L.O.C we have \[ BX^2 = BF^2 + FX^2 - 2BF \cdot FX \cos(\angle BFX). \] Substituting, we obtain \[ BX^2 = (s-b)^2 + (s-a)^2 \sin^2 \frac{\alpha}{2} - 2(s-b)(s-a)(\sin \frac{\alpha}{2})(-\sin \frac{\alpha}{2}). \] From here, we find \[ BX^2 = (s-b)^2 + (s-a)^2 \sin^2 \frac{\alpha}{2} + 2(s-b)(s-a)(\sin^2 \frac{\alpha}{2}) = (s-b)^2 + (3s-a-2b)(s-a)(\sin^2 \frac{\alpha}{2}). \] Thus, we know \[ BX^2 = (s-b)^2 + (s+c-b)(s-a)(\sin^2 \frac{\alpha}{2}). \] By symmetry, we also have \[ CX^2 = (s-c)^2 + (s+b-c)(s-a)(\sin^2 \frac{\alpha}{2}). \] Combining these, we find \[ BX^2 + CX^2 = (s-b)^2 + (s-c)^2 + 2s(s-a)(\sin^2 \frac{\alpha}{2}). \] However, note that \[ a^2 = ([s-b] + [s-c])^2 = (s-b)^2 + (s-c)^2 + 2(s-b)(s-c). \] To prove \(a^2 > BX^2 + CX^2\), it suffices to show that \(a^2 - BX^2 - CX^2 > 0\). By subtracting the two expressions, we obtain \[ a^2 - BX^2 - CX^2 = 2(s-b)(s-c) - 2s(s-a)(\sin^2 \frac{\alpha}{2}). \] Note that \[ \sin^2 \frac{\alpha}{2} = \frac{1 - \cos \alpha}{2}. \] Thus, it is sufficient to prove \[ 2(s-b)(s-c) > s(s-a)(1 - \cos \alpha). \] Using L.O.C, we find \[ \cos \alpha = \frac{b^2 + c^2 - a^2}{2bc}. \] Substituting, we have \[ s(s-a)(1 - \cos \alpha) = s(s-a)\left(1 - \frac{b^2 + c^2 - a^2}{2bc}\right) = s(s-a)\left(\frac{a^2 - (b-c)^2}{2bc}\right). \] Hence, substituting this into the inequality, it suffices to prove \[ 2(s-b)(s-c) > s(s-a)\left(\frac{a^2 - (b-c)^2}{2bc}\right). \] Using Ravi’s substitution, where \(x, y, z \in \mathbb{R^+}\) are defined as \(a = x + y\), \(b = y + z\), and \(c = z + x\), we substitute these into the inequality. It then suffices to prove \[ 2xy > z(x+y+z)\left(\frac{(x+y)^2 - (x-y)^2}{2(y+z)(z+x)}\right) = z(x+y+z)\frac{2xy}{(y+z)(z+x)}. \] Thus, it suffices to prove \[ (y+z)(x+z) > z(x+y+z). \] This is clearly true because \(xy > 0\).