+830 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.
+195 Since M is the midpoint of EF and N is the midpoint of EH, we have FM=ME=EN=NH. This means triangles FMP and HNP are similar.
Furthermore, since EFGH is a parallelogram, then ∠FPH=∠HPN (alternate interior angles).
Therefore, triangles FPH and HNP are also similar by Angle-Angle Similarity (AA).
Since the triangles are similar, we have the following proportion:
NPPH=HPFP
Cross-multiplying gives us PH2=FP⋅NP. However, we want to find PQ/FH. Since FMP and HNP are similar, we also have:
FMPQ=PHFP
Cross-multiplying again gives us PQ⋅PH=FM⋅FP. Substituting the first equation we found (PH2=FP⋅NP) into this second equation, we get:
PQ⋅PH=FM⋅FP=FM⋅PH2
=FM⋅PH
Dividing both sides by PH (which is nonzero since P is on line segment FH), we get PQ=FM. Finally, since FM=ME=21EF, we have:
FHPQ=FHFM=2FHEF=21
