Prove Midpoints of Quadrilateral Form Parallelogram

[spoc21]

Homework Statement

Use Cartesian vectors in two-space to prove that the line segments joining midpoints of the consecutive sides of a quadrilateral form a parallelogram.

Homework Equations

We could say vectors a and b

a = [a₁, a₂]
b = [b₁, b₂]

The Attempt at a Solution

Am really confused by this question..I would really appreciate it if anyone could offer tips to help me get started on this problem.

Thanks!

 

[tiny-tim]

Hi spoc21!

We could say vectors a and b

a = [a₁, a₂]
b = [b₁, b₂]

So far so good!

Next would be

c = [c₁, c₂]
d = [d₁, d₂]

Now what are the midpoints of ab, bc, cd, and da ?

 

[spoc21]

Hi tiny-tim,

so if we use the midpoint formula, we would get the following:

midpoint of ab = (a₁+b₁/2, a₂ + b₂/2)

midpoint of bc = (b₁ + c₁/2, b₂c₂/2)

midpoint of cd = ( c₁+ d₁/2, c₂ + d₂/2)

midpoint of da = (d₁ + a₁/2, d₂ + a₂/2)

Now I'm stuck, how would I show that joining these points would produce a parallelogram.

Thank you very much for your help!

 

[Mark44]

You should identify the midpoints - give them names, such as by saying the E is the midpoint between A and B, F is the midpoint between B and C, G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A.

Now if the segments EF and GH are parallel, and the segments FG and HI are parallel, then EFGH is a parallelogram.

One nit: In your midpoint formulas, add parentheses around the sums. E.g., ((a1 + b1)/2, (a2 + b2)/2)

 

[spoc21]

ok, so I get the following results

EF = [(c₁ - a₁)/2, (c₂ - a₂)/2]
GH = [(a₁ - c₁)/2, a₂ - c₂)/2]

I'm a little lost here..(don't know how to proceed)

Also, you said "G is the midpoint between C and D, H is the midpoint between C and D, and I is the midpoint between D and A."

Are there two midpoints G, as well as H between C, and D?

Thanks for your help

 

[Mark44]

I lost track of my midpoints. There should only be four of them E, F, G, and H - no I needed. I meant "H is the midpoint between D and A."

You can check that your vectors EF and GH are parallel by showing that their slopes are the same. The slope of a vector v = <a, b> is b/a.

 

[spoc21]

ok thanks Mark..also, I was wondering if there is a way to solve this using the properties of cross products?

 

[Mark44]

|u X v| = |u||v|sin(theta), where theta is the angle between u and v. If u and v are parallel, the angle between them will be 0 or pi, so the magnitude of the cross product will be zero.

 

[spoc21]

So I have attached my working in pdf format, and would appreciate if you could just take a quick look at it, as I am unsure if this is correct...Thanks!

 

[tiny-tim]

Hi spoc21!

ok, but a little long-winded.

(and why are you using square brackets instead of the usual round ones? has your professor told you to? )

It would have saved time and space to choose an easier notation …

let A = (x_A,y_A) etc

then the midpoints are M_AB = 1/2 (x_B + x_A, y_B + y_A) etc

and the vectors joining consecutive midpoints are V_AC = 1/2 (x_C - x_A, y_C - y_A) etc

so V_AC = -V_CA and V_BD = -V_DB

in other words, opposite sides of the quadrilateral are parallel, and so it must be a parallelogram ("equal in magnitude" is a bonus, but not necessary for the proof)

(note incidentally that there would have been no need to use Cartesian coordinates if the question hadn't required them … you can just use vector symbols, eg m_AB = 1/2 (b + a) etc )

EDIT: just noticed another way, that throws slightly more light on the position of the parallelogram …

use the fact that if the midpoints of the two diagonals of a quadrilateral coincide, then it's a parallelogram …

and those midpoints, for the new quadrilateral, are (both) 1/4 (a + b + c + d) …

so it's a parallelogram, and "centred" in the same place as the original quadrilateral ​