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.

magenta Apr 15, 2024

#1**0 **

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

Boseo Apr 15, 2024