22nd Chinese Girls' Mathematical Olympiad

Proof: On the circumcircle of $△ABC$, a point $F$ is chosen such that $ABCF$ forms an isosceles trapezoid. The line $FH$ intersects the circumcircle of $△ABC$ at another point $L$, and intersects the median $AK$ at point $G^{\prime }$. As shown in the figure. Since $AF=2HK$, it follows that $\frac{A{G}^{\prime }}{G^{\prime }K}=\frac{AF}{HK}=2$, implying that $G^{\prime }$ is the centroid of $△ABC$, hence $G^{\prime }=G$. 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 $△ABC$ 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. $\square$