How do I prove that the midpoint of the sides of a quadrilateral form a parallelogram?

Answer 1

You take any 4 points
(x1,y1)
(x2,y2)
(x3,y3)
(x4,y4)

The 4th point can be shifted to the origin (0,0)
The coordinates of the graph can be rotated and expanded so that the 3rd point is at (0,1)

So your 4 points are now
(x1,y1)
(x2,y2)
(0,1)
(0,0)

find the mid-point of points1,2
2,3
3,4
4,0

( (x1+x2)/2,(y1+y2)/2 )
(x2/2, (y2+1)/2 )
(0,1/2)
( x1/2, y1/2)

Find the slope between each mid-point. Each of these slopes is the slope of a side of the parallelogram.
(y1-1)/x1
y2/x2
(y1-1)/x1
y2/x2

The first and third slope are the same(which means the sides are parallel.

And the second and fourth slopes are the same

Answer 2

You can prove that he midpoint of the sides of a "parallelogram" form a parallelogram.

A parallelogram's opposite sides are of equal length, and it's opposite angles are of the same measurement.

Use this information, and prove that when joining midpoints of adjacent sides, the opposite triangles formed are congruent(SAS).

Then you will be able to prove that the new quadrilateral have two pairs of equal opposite sides, which makes it a parallelogram.

Answer 3

Suppose that ABCD is the quadrilateral, and K, L, M, N are the midpoints of sides AB, BC, CD, DA. Draw lines KN, LM and the diagonal BD.

Now, the line segment that connects the midpoints of the two sides of a triangle is parallel to the third side and it is equal to 1/2 of the third side. So from the triangle ABD:

KN = BD/2 and KN is parallel to BD

and from the triangle BCD

ML = BD/2 and ML is parallel to BD

At last KN = ML and KN is parallel to BD, so KLMN is a parallelogram

Answer 4

hmmm...here I go...

STATEMENT
1. Quadrilateral ACEG
2. B is midpoint of AC
3. D is midpoint of CE
4. F is midpoint of EG
5. H is midpoint of AG
6. draw auxiliary line AE
7. draw auxiliary line BD
8. BD is a median of triangle ACE
9. BD is parallel to AE
10. draw auxiliary line HF
11. HF is a median of triangle AGE
12. HF is parallel to AE
13. BD is parallel to HF

REASON
1. given
2. given
3. given
4. given
5. given
6. auxiliary (extra) line
7. auxiliary line
8. B and D are midpoints
9. definition of median
10. auxiliary line
11. H and F are midpoints
12. definition of median
13. lines parallel to the same line are parallel to each other

...and you do the same thing for BH and DF and then prove that BDFH is a parallelogram by definition of a parallelogram...
The only thing I'm not sure of though are numbers 8 and 11...I'm not sure if it's called a median or something else...but it's that segment in a triangle where if the sides are proportional, that segment is parallel to the base...I dunno...but you get the idea...lol

Hope this helped!

Answer 5

Draw a quadrilateral like so: starting up from an arbitrarty aspect "O", draw 2 vectors "a" and "d", in 2 diverse instructions. next from the authentic of "a", draw yet another vector "b" oftentimes in route of the endpoint of "d". finally, connect the endpoint of "d" to the endpoint of "b" with yet another vector "c". If I defined it proper, you need to have a 4 sided make certain with vectors "a" & "b" terminating on an analogous aspect as vectors "c" & "d". Now, be conscious that there are 2 diagonals. enable's call the only connecting aspect "O" to the different nook "y", and the different one "x". we may be able to make certain that y = a + b = d + c x = a - d = b - c next, we see that the vector connecting the midpoints of "a" and "b" is a million/2(a+b). it is so because the area vector to at least one end is a million/2a and to the different is a + a million/2b, and subtracting supplies (a + a million/2b) - a million/2a = a million/2a + a million/2b = a million/2(a+b) with the help of similar reasoning we get the vector connecting mid-b to mid-c a million/2(b-c) and from mid-a to mid-d a million/2(a-d) and finally from mid-d to mid-c a million/2(d+c) Reviewing our diagonals {x,y} we may be able to make certain that a million/2(a+b) = a million/2(d+c) and a million/2(b-c) = a million/2(a-d) and in view that the vectors on opposite sides of the make certain connecting midpoints are equivalent, that make certain is a parallogram. QED

Answer 6

that is not true because, for instance if you use a concave quadrilateral (like an arrow head shape) this does not work. Good luck.

Answer 7

i didn't think u could do that :D