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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.

 Jun 4, 2024
 #1
avatar+129399 
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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

 

cool cool cool

 Jun 4, 2024

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