+436 In parallelogram $EFGH,$ let $M$ be the point on $\overline{EF}$ such that $FM:ME = 1:1,$ and let $N$ be the point on $\overline{EH}$ such that $HN:NE = 1:1.$ Line segments $\overline{FH}$ and $\overline{GM}$ intersect at $P,$ and line segments $\overline{FH}$ and GN intersect at Q. Find PQ/FH.
+129771 
Triangle HQN is similar to triangle FQG Triangle HPG is similar to triangle FPM
HQ / NH = FQ / GF PH / HG = PF / FM
HQ / 2 = FQ / 4 PH / 4 = PF / 2
HQ / FQ = 2/4 = 1/2 PF / PH = 2/4 =1/2
HQ = (1/3)FH
PF = (1/3) FH
PQ = (FH - PF - HQ) = (FH - (1/3)FH - (1/3)FH) = (1/3)FH
So
PQ / FH = 1/3
