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I don't know what to do here!!!!  Help!!!

 

In parallelogram $EFGH,$ let $M$ be the point on $\overline{EF}$ such that $FM:ME = 3:5,$ and let $N$ be the point on $\overline{EH}$ such that $HN:NE = 2:5.$  Line segments $\overline{FH}$ and $\overline{GM}$ intersect at $P,$ and line segments $\overline{FH}$ and $\overline{GN}$ intersect at $Q.$  Find $\frac{PQ}{FH}.$

 Dec 14, 2023
 #1
avatar+129270 
+1

 

 

 

Triangle HQN similar to Triangle FQG

HN / FG =  2 / 7

Therefore  HQ / FQ  = 2 / 7

Therefore  HQ / (HQ + FQ) =  HQ / HF  =  2 / ( 2 + 7)  = 2/9  ⇒ HQ = (2/9) HF

 

Triangle PFM similar to Triangle PHG

FM / HG =  3 / 8

Therefore PF / PH = 3 / 8

Therfore PF / (PF + PH)  = PF / HF  = 3 /( 3 + 8) = 3 / 11 ⇒  PF = (3/11) HF 

 

Getting a common denominator  between  9  and 11

HQ = 22/99FH

PF  = 27/99 FH

 

Therefore

FH - HQ - PF  =  PQ

FH - (22/99)FH - (27/99)FH  = PQ

(50/99)FH  = PQ

 

Therefore

PQ / FH  =  50 / 99

 

cool cool cool

 Dec 14, 2023

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