Prove: In any quadrilateral ABCD. Connect the consecutive mid points of each line.?

Question

The figure formed is a parallelogram.

recall the definition of a median of a triangle
median of a triangle is line that is parallel to the base and is equal to one-half of it

scythian proved that the opposite lines are equal but does not prove they are parallel

Mr. Lee proved that the opposite lines are parallel but did not prove that they are equal

I will just vote one on each of you

Answer 1

Here is a ROUGH sketch for reference, with vertices A, B, C, and D, with midpoints w, x, y, and z...
A------y______B
.|.....................|
x|....................|z
...|...................|
..D------w____C
To do this proof, you need to utilize the triangle property that states:
The segment formed by the midpoints of two adjacent sides is parallel to the third side and is half the length of the third side.
The only part of it you need though is the parallel portion of the statement. Now,
1) Construct diagonal BD. 1) Definition of diagonal/two points determine a line.
2) yx || BD........................... 2) Triangle property from above.
3) zw || BD...........................3) Triangle property.
4) yx || zw.............................4) Transitive law of parallel lines?
(I'm not entirely sure of the exact reason, but you can deduce that if two lines are parallel to a third line, then the first two lines are parallel). Now do this again with the other diagonal.
5) Construct diagonal AC. 5) Definition of diagonal/...
6) xw || AC...........................6) Triangle prop.
7) yz || AC............................7) Triangle prop.
8) xw || yz.............................8) Transitive law of || lines.
Now, since you have opposite sides of a quadrilateral are parallel, this by definition is a parallellogram.
9) wxyz is a parallelogram. 9) Definition of parallelogram.
There, you're done. Hope this helped.

Answer 2

Let vectors A, B, C, D be the corners of this quadrilateral. Then the midpoints are (1/2)(A+B), (1/2)(B+C), (1/2)(C+D), (1/2)(D+A). We can take differences to find the vector sides of the inscribed quadrilateral: (1/2)(A+B-B-C), (1/2)(B+C-C-D), (1/2)(C+D-D-A), (1/2)(D+A-A-B), or (1/2)(A-C), (1/2)(B-D), (1/2)(C-A), (1/2)(D-B). A quick inspection shows that we have two pairs of equal vectors, but for parity. QED.

Answer 3

a square