Sketch: It is enough to prove $PQ||BC$, $MP||AB$, $MQ||AC$. By Ceva $AM$, $BY$,$CX$ concur - denote this point by $N$. From this we get by reversed Ceva (for point $N$ in triangle $GBC$) and Thales $PQ||BC$. Call intersetion of $CG$ and $BG$ with sides $C'$ and $B'$ respectively. By Menelaos in CC'X and line GB we have $3QX=CQ$ and by Thales $3GP=PC$. Now: $\frac{C'P}{PC}=\frac{C'G+GP}{PC}=\frac{1}{3}+\frac{CG}{2PC}=\frac{1}{3}+\frac{GP+PC}{2PC}=1$ Hence $MP||AB$ and similarly $MQ||AC$. QED

Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=43338 . darij

Since $XY||BC$, then from Ceva, it follows that $AM, BY$ and $CX$ are concurrent. Let $\alpha=\angle{BCP}$ and $\beta=\angle{ACP}$. Then from the Sine Theorem, $\frac{BP}{\sin\alpha}=\frac{BC}{\sin(\angle{BPC})}$ and $\frac{PY}{\sin\beta}=\frac{YC}{\sin(\angle{YPC})}=\frac{\frac{1}{3}AC}{\sin(\angle{BPC})}$. So $\frac{BP}{PY}=\frac{BC\sin\alpha}{\frac{1}{3}\sin\beta}$. Denote $C_{1}$ by the midpoint of $AB$. Using similar arguments, $\frac{BC_{1}}{C_{1}A}=1=\frac{BC\sin\alpha}{AC\sin\beta}$. So $\frac{BP}{PY}=3$. From Ceva for $\triangle{BGC}$ it results that $QP||BC||XY$. So $\frac{BQ}{QG}=3$. Let $B_{1}$ be the midpoint of $AC$. It results that $Q$ is the midpoint of $BB_{1}$. So $QM||AC$. Using similar arguments $PM||AB$. So $\triangle{ABC}$ and $\triangle{MPQ}$ have parallel sides. So they're similar.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1991Problem1 Vo Duc Dien

suppose $CG$ intersect $AB$ at $R$ and $BG$ intersect $AC$ at $S$ : lemma 1:$M$and$Q$and$R$ are on same line. proof of lemma 1: on triangle $GBC$ we must proof: $\frac{RG}{RC} . \frac{CM}{MB} . \frac{BQ}{QG} =1 $ so we must proof $\frac{BQ}{QG} = 3$ we know that : $\frac{BQ}{QG} = \frac{CB}{GC} . \frac{\sin QCB}{\sin QCG} \rightarrow \frac{BQ}{QG}= \frac{CB}{GC} . \frac{GC}{GX} = 3$ end of proof of lemma1. proof of problem: so $QMP=\angle A$ by menlaues on triangle $RMC$ and point $B,Q,G$ and also $SMB$ and point $C,P,G$ so we know that: $\frac{MQ}{AC} = \frac{MP}{AB}$ end of proof of problem. oh excuse me but im so lazy and this solution is so short!

Let $\Delta MM_BM_C$ be the medial triangle WRT $\Delta ABC$ and then apply Ceva on $\Delta ABC$, $\Delta GBC$, $\Delta BM_CC$ and $\Delta BM_BC$ to obtain $\Delta MPQ$ $\sim$ $\Delta ABC$ @below I didn't expect a guy like you to bash such a beautiful geo

This shows the power of bary As usual take $A,B,C=(1,0,0),(0,1,0),(0,0,1)$.Then obtain $M=(0,\dfrac{1}{2},\dfrac{1}{2}),X=(\dfrac{1}{3},\dfrac{2}{3},0),Y=(\dfrac{1}{3},0,\dfrac{2}{3}),G=(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}),P=(\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{2}),Q=(\dfrac{1}{4},\dfrac{1}{2},\dfrac{1}{4})$.We are ready for computation.Since $|PQ|^2=-a^2yz-b^2xz-c^2xy$ where $PQ$ is the displacement vector, so its easy to see $|PQ|^2=\dfrac{a^2}{16},|MQ|^2=\dfrac{b^2}{16}$.Thus $MPQ\sim ABC$ with similarity factor $\dfrac{1}{4}$.This center of similarity is $AM\cap BP \cap CQ$.

Surprised this solution hasn't been posted yet! Maybe it's fundamentally similar to some of the others. First remark by similarity ratios that $\tfrac{PC}{PG} = \tfrac{BQ}{QG} = 2$, so $PQ\parallel BC$. Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Triangles $EGY$ and $EBC$ are homothetic, as are triangles $PGY$ and $PCB$; looking at both centers of homothety shows that $E$, $P$, and $M$ are collinear. Likewise, $D$, $Q$, and $M$ are collinear. Therefore $\triangle ABC\sim\triangle MED\sim\triangle MPQ$.

Denote the midpoint of $\overline{AC}$ and $\overline{AB}$ as $D$ and $E$. Notice that $GY = \tfrac{1}{3} BC$, so $\triangle GPY \sim \triangle BPC$ with $\tfrac{GP}{PC} = \frac{1}{3}$. Hence, $PE=PC$, so $\triangle CPM \sim \triangle CEB$ which implies $\overline{PM} \parallel \overline{AB}$. Similarly, $\overline{QM} \parallel \overline{AC}$, which implies the desired similarity. $\square$