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In parallelogram EFGH, let M be the midpoint of side \(\overline{EF}\), and let N be the midpoint 

of side \(\overline{EH}\). Line segments \(\overline{FH}\) and \(\overline{GM}\) intersect at P, and line segments \(\overline{FH}\) and \(\overline{GN}\) intersect at Q. Find \(\frac{PQ}{FH}\).

 May 1, 2020
 #1
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See the following image :

 

 

Let E  = (2,4)    F  = (6,4)     G = (4,0)     H = (0,0)

 

FH  =  sqrt  ( 6^2 + 4^2)  = sqrt ( 36 + 16)  = sqrt (52)   

 

Midpoint of EF  =  (4,4)   = M

Midpoint of  EH = (1,2)  = N

 

Equation of line  through MG  is  x  =  4

Equation of line through FH is  y = (4/6)x  = (2/3)x

The x coordinate of P  = 4

The y coordinate of P  is  y = (2/3)(4)  = 8/3

So P =  (4, 8/3)

 

Similarly

The slope  of the line through GN  is ( 2-0) / (1 - 4)   = -2/3

And the equation of of this line is   y = (-2/3)(x - 4)  = (-2/3)x + 8/3

 

The find the x coordinate of Q

(-2/3)x + 8/3  = (2/3)x

8/3 = (4/3)x

x = 2

And the y coordinate of Q  is    (2/3)(2)  = 4/3

So Q  = ( 2, 4/3)

And PQ  =  sqrt  [( 4-2)^2  + (8/3 - 4/3)^2]  =  sqrt  [ 2^2 + (4/3)^2]  = sqrt [ 4 + 16/9]  =

sqrt [36 + 16] / 3  =   sqrt (52)/3

 

So

 

PQ                  sqrt(52) / 3                 1

___      =         __________   =      ____

FH                   sqrt (52)                     3

 

 

cool cool cool

 May 1, 2020

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