We proceed with barycentrics. Let one line be $ax+by+cz = 0$, WLOG $a>b>c$ as if two parameters are equal, the line will not intersect the edge opposite the third vertex, then the other line must pass through $(0, b, -c), (a, -b, 0), (a, 0, -c)$ and must be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 0$. The intersection is thus $\left[-\frac{a(b+c)}{(a-b)(a-c)}\right]$, which is homogenous. The medial triangle requirement is equivalent to saying that the point lies within the triangle after $\mathcal{H}(G, -2)$, giving $3G-2P$, hence the condition is equivalent to a cyclic truth of $\frac{(a+b)(a+c)}{(a-b)(a-c)} > 0$. Multiplying all three of these parameters, we have $-\left(\frac{(a+b)(b+c)(c+a)}{(a-b)(b-c)(c-a)}\right)^2 > 0$, which is a contradiction, and we are done.

@above nice! Here is a synthetic or I should rather say a projective and affine approach to this problem. Let $\ell,\ell^*$ be two isotomic lines WRT $\Delta ABC$. $\ell\cap AB,AC=D,E$ and $\ell^*\cap AB,AC=D^*,E^*$. Keep $E$ fixed and animate $D$ over $\overline{BC}$. Hence, the map $\ell\mapsto \ell^*$ is a homography and $\ell\cancel{\equiv} \ell^*$ on any input of $D$. Hence, $\ell\cap\ell^*$ traces out a conic through $X,Y,E,E^*$ and $\infty_{XZ},\infty_{YZ}$ (so it must be a hyperbola as it has two points at infinity) where $\Delta XYZ$ is the medial triangle of $XYZ$. So it boils down to prove the fact (this is easy using cart bash but I'm looking for a synthetic one) that

Any point on the hyperbola passing through $B,C,\infty_{AB},\infty_{AC}$ of a triangle $\Delta ABC$ cannot lie inside $\Delta ABC$. Further

Using Menelaus's theorem, we can solve the problem easily.

meysam1371 wrote: Using Menelaus's theorem, we can solve the problem easily. How?