Proof: On the circumcircle of $\triangle ABC$, a point $F$ is chosen such that $ABCF$ forms an isosceles trapezoid. The line $FH$ intersects the circumcircle of $\triangle ABC$ at another point $L$, and intersects the median $AK$ at point $G'$. As shown in the figure.
Since $AF = 2HK$, it follows that $\frac{AG'}{G'K} = \frac{AF}{HK} = 2$, implying that $G'$ is the centroid of $\triangle ABC$, hence $G' = G$. Therefore,
$$\angle BLH = \angle BLF = \frac{1}{2}\widehat{BF} = \frac{1}{2}\widehat{AC} = \angle ABC,$$
Given $\angle BDH + \angle ABC = 180^\circ$, it follows that $\angle BDH + \angle BLH = 180^\circ$, which means points $B, L, H, D$ are concyclic.
Similarly, it can be proved that $\angle CLH = \angle ACB$, and points $C, L, H, E$ are concyclic.
Since $\angle BLH = \angle ABH$, $PB$ is tangent to circle $\odot BLHD$ at point $B$. Let the incircle of $\triangle ABC$ be $\omega$, and $PB$ is an external common tangent of $\omega$ and $\odot BLHD$. Given $MP = MB$, it is known that $M$ is the radical center of $\omega$ and $\odot BLHD$, thus line $MD$ is the radical axis of $\omega$ and $\odot BLHD$.
Similarly, it can be proved that line $NE$ is the radical axis of $\omega$ and $\odot CLHE$. Moreover, line $GH$ is the radical axis of $\odot BLHD$ and $\odot CLHE$, therefore lines $MD, NE, GH$ either intersect at a single point, or are pairwise parallel.
If $MD, NE, GH$ are pairwise parallel, then the centers $O_1, O_2$ of $\odot BLHD, \odot CLHE$, and the center $I$ of $\omega$ are collinear. Since both $\angle BDH$ and $\angle CEH$ are obtuse, $O_1, O_2$ are below $BC$, obviously $I$ is above $BC$. Let the projections of $O_1, O_2,$

$I$ on $BC$ be $X, Y, Z$ respectively, then $X, Y$ are the midpoints of $BH, CH$ respectively. Given $AB < AC$, it is known that $Y, Z$ are on the same side of $AH$, and
$$CZ = \frac{AC + BC - AB}{2} > \frac{BC}{2} > \frac{CH}{2} = CY.$$
Hence, $Z$ lies on the segment $XY$, therefore $O_1, O_2, I$ cannot be collinear, a contradiction. Thus, $MD, NE, GH$ intersect at a single point. $\square$