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 \(\frac{PQ}{FH}\).
+128474 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
