Proving a Quadrilateral is a Parallelogram

  • Thread starter Thread starter lovemake1
  • Start date Start date
  • Tags Tags
    Planes
AI Thread Summary
The discussion focuses on proving that the quadrilateral formed by connecting the midpoints of a given quadrilateral's sides is a parallelogram. Participants clarify that the original quadrilateral does not need to be a parallelogram for this to hold true. To prove this, one must find the midpoints of the sides, calculate the slopes of the segments connecting these midpoints, and show that opposite sides have equal slopes. A misunderstanding arises regarding the definition of adjacent sides, which are clarified as touching sides. Ultimately, the correct method involves confirming that the slopes of both pairs of opposite sides are equal, establishing the figure as a parallelogram.
lovemake1
Messages
147
Reaction score
1

Homework Statement



p1(x1,y1) , p2(x2,y2), p3(x3,y3), p4(x4,y4) are the vertices of a quadrilateral. Show that the quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram.

Homework Equations



Midpoint = x1 + x2 / 2 and y1 + y2 / 2


The Attempt at a Solution



im not quite sure of what the question is asking...
"show that quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram"
isnt this just mean cut the original paralleogram in half ? and still proove its a prallelogram ?

how can a trapezoid for example turn into parallelogram after joining the midpoints of adjacent sides... right it cant..
This is why the question is so misleading to me..

can anyone give me a small clue? so i can try to figure this question out by myself thanks !
 
Physics news on Phys.org
lovemake1 said:

Homework Statement



p1(x1,y1) , p2(x2,y2), p3(x3,y3), p4(x4,y4) are the vertices of a quadrilateral. Show that the quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram.

Homework Equations



Midpoint = x1 + x2 / 2 and y1 + y2 / 2


The Attempt at a Solution



im not quite sure of what the question is asking...
"show that quadrilateral formed by joining the midpoints of adjacent sides is a parallelogram"
isnt this just mean cut the original paralleogram in half ?
No, you're given a quadrilateral - it's not give that it's a parallelogram. The question is saying that if you connect the midpoints of any four-sided figure, what you get is a parallelogram
lovemake1 said:
and still proove its a prallelogram ?

how can a trapezoid for example turn into parallelogram after joining the midpoints of adjacent sides... right it cant..
No, not right. The resulting figure is a parallelogram, and that's what you need to prove, except you can't assume you're starting with a trapezoid (or rectangle etc).
lovemake1 said:
This is why the question is so misleading to me..

can anyone give me a small clue? so i can try to figure this question out by myself thanks !
Find the midpoints of all four sides.
Calculate the slopes of the opposite sides of the figure you get. If the slopes are equal, you have a parallelogram.
 
hmm. so you mean like a form a diamond within the 4 sides right ? i was more of thinking adjacent sides meaning one side that's not touching the other.
so like one across from another, making an X shape.

so if we were to form a diamond, how would we use 4 points to find the slope?
i know that y1 - y0 / x1 - x0 is slope for one side. But that doesn't give me any clue to if the sides have same slope.
 
i've got all 4 Mid points of each sides.
just using the variables P1x, P2x, P3x, P4x, Py1, Py2, Py3, Py4.

Each Midpoint looks like this. Midpoint-1: (P1x + P2x) / 2 , (P1y + P2y) / 2
after getting 4 mid points, I've tried to get slope from using these points.

They looked as follows: Slope of midpoint 1 and 4 = [(p1x + p4x) - (p2x + p3x) / 2] / [(p1y + p4y - p2y + p3y)/2]

am i on the right track here?

i can compare this with slope of midpoint 2 and 3 which is opposite of midpoint 1 and 4.
For some reason i did not get the same variable.

am i doing this completely wrong ? please help
 
lovemake1 said:
hmm. so you mean like a form a diamond within the 4 sides right ?
Yes, only it's not a diamond. You're proving that it's a parallelogram.
lovemake1 said:
i was more of thinking adjacent sides meaning one side that's not touching the other.
Adjacent sides do touch (meet). Opposites sides don't meet.
lovemake1 said:
so like one across from another, making an X shape.

so if we were to form a diamond, how would we use 4 points to find the slope?
i know that y1 - y0 / x1 - x0 is slope for one side. But that doesn't give me any clue to if the sides have same slope.

Compare (y1 - y0) /(x1 - x0) for one segment with the slope of the segement across from it. If you're not drawing a picture of this, you should be.
 
lovemake1 said:
i've got all 4 Mid points of each sides.
just using the variables P1x, P2x, P3x, P4x, Py1, Py2, Py3, Py4.

Each Midpoint looks like this. Midpoint-1: (P1x + P2x) / 2 , (P1y + P2y) / 2
after getting 4 mid points, I've tried to get slope from using these points.

They looked as follows: Slope of midpoint 1 and 4 = [(p1x + p4x) - (p2x + p3x) / 2] / [(p1y + p4y - p2y + p3y)/2]

am i on the right track here?
Maybe. It's hard to say for sure without seeing a picture. If your four midpoints are M1, M2, M3, and M4 (going around the figure, clockwise or counterclockwise), you want to compare the slope of the segment from M1 to M2, with the slope of the segment from M3 to M4, and do the same for the segment from M2 to M3 compared to M4 to M1.
lovemake1 said:
i can compare this with slope of midpoint 2 and 3 which is opposite of midpoint 1 and 4.
For some reason i did not get the same variable.

am i doing this completely wrong ? please help
 
NVM. i had the right answer in the first place.
i was very stupid. didnt sub in the right variables :D
Thanks, problem solved !
 
I haven't worked it out, but I think it should work. Assuming that M1, M2, M3, and M4 are the midpoints of the original quadrilateral, and are labelled going around, what do you have for the slope from M1 to M2 and for M3 to M4?
 
m1 to m2 : p2y + p3y - (p1y + p2y) / p2x + p3x - (p1x + p2x )

Reduced to : p3y - p1y / p3x - p1y

m4 to m3 : p3y + p4y - ( p1y + p4y ) / p3x + p4x - ( p1x + p4x )

Reduced to : p3y - p1y / p3x - p1x

P.S- I removed the division of 2 from the midpoint equation. because they cancle out.

Got this answer. This is the correct method to achieve the solution right ?
The Slope of the two lines seems to match.
 
  • #10
If the slopes of two opposite sides are equal, that's what you wanted to show. Now show that the slopes of the other two opposite sides are equal, and you'll be done.
 
Back
Top