The vertices of quadrilateral DEFG are D(3,2) E(7,4) F(9,8) and G(5,6).?

Question

The vertices of quadrilateral DEFG are D(3,2) E(7,4) F(9,8) and G(5,6).

a) Prove DF and GE bisect each other
b) DEFG is a rhombus

Answer 1

a) Find the point of intersection of DF and GE. Call that point M.
Show that the length of DM = MF and the length of GM = ME.
b) Find the length between each set of points (DE, EF, FG and GD). If they are all equal, then it is a rhombus.

Answer 2

Not certain if this is true, since it has been awhile, but

if (DF)^2 + (GE)^2 = 2((DE)^2 + (EF)^2), then the 2 lines bisect

(sqrt(72))^2 + (sqrt(8))^2 = 2((sqrt(20))^2 + (sqrt(20))^2)
72 + 8 = 2(20 + 20)
80 = 2(40)
80 = 80

Therefore the 2 lines bisect.

Sorry if this isn't correct.

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b.)
(3,2) and (7,4)
D = sqrt((7 - 3)^2 + (4 - 2)^2)
D = sqrt(4^2 + 2^2)
D = sqrt(16 + 4)
D = sqrt(20)

(7,4) and (9,8)
D = sqrt((9 - 7)^2 + (8 - 4)^2)
D = sqrt(2^2 + 4^2)
D = sqrt(4 + 16)
D = sqrt(20)

(9,8) and (5,6)
D = sqrt((5 - 9)^2 + (6 - 8)^2)
D = sqrt((-4)^2 + (-2)^2)
D = sqrt(16 + 4)
D = sqrt(20)

(3,2) and (5,6)
D = sqrt((5 - 3)^2 + (6 - 2)^2)
D = sqrt(2^2 + 4^2)
D = sqrt(4 + 16)
D = 20

As you can see all sides equal, so that means its either a square or a rhombus

If DF does not equal EG, then it can only be a rhombus

(3,2) and (9,8)
D = sqrt((9 - 3)^2 + (8 - 2)^2)
D = sqrt(6^2 + 6^2)
D = sqrt(36 + 36)
D = sqrt(72)

(7,4) and (5,6)
D = sqrt((5 - 7)^2 + (6 - 4)^2)
D = sqrt((-2)^2 + 2^2)
D = sqrt(4 + 4)
D = sqrt(8)

Answer 3

hmmm

a) Show that the midpoint of DF and GE are the same point and then you'll have it.

b) Show that all 4 segments of the quad are of equal length and it's a rhombus.