So, initially 240
![\frac{3}{8}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B8%7D%20)
-212
![\frac{12}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B12%7D%7B3%7D%20)
were empty:
this is :
![240 \frac{3}{8} - 212 \frac{12}{3} = 240 \frac{3}{8} - 212 +4 = 240 \frac{3}{8} - 216= 24 \frac{3}{8}](https://tex.z-dn.net/?f=240%20%20%5Cfrac%7B3%7D%7B8%7D%20-%20212%20%20%5Cfrac%7B12%7D%7B3%7D%20%20%3D%20240%20%20%5Cfrac%7B3%7D%7B8%7D%20-%20212%20%20%2B4%20%3D%20240%20%20%5Cfrac%7B3%7D%7B8%7D%20-%20216%3D%2024%20%20%5Cfrac%7B3%7D%7B8%7D%20)
12/4 is 4 so that's why i substituted it with 4.
now, later 27 1/3 were used so we add this tho the original empty space
![24 \frac{3}{8}+ 27 \frac{1}{3}](https://tex.z-dn.net/?f=%2024%20%20%5Cfrac%7B3%7D%7B8%7D%2B%2027%20%20%5Cfrac%7B1%7D%7B3%7D)
=
![24 \frac{9}{24}+ 27 \frac{8}{24}](https://tex.z-dn.net/?f=%2024%20%20%5Cfrac%7B9%7D%7B24%7D%2B%2027%20%20%5Cfrac%7B8%7D%7B24%7D)
=
which is the result!
Step 1
The property of a parallelogram is that diagonals of a parallelogram bisect each other and hence have the same midpoint.
This means that the diagonals cut themselves into two equal halves.
Hence, they must have the same mid-points.
Therefore, if we prove that AC and BD have the same midpoint, then ABCD is a parallelogram.
Step 2
Find the midpoint of BD and AC
![\begin{gathered} (\frac{x_2+x_1}{2}),(\frac{y_2+y_1}{2}) \\ BD=(-\frac{1}{2},\frac{9}{2}) \\ AC=(-\frac{1}{2},\frac{9}{2}) \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28%5Cfrac%7Bx_2%2Bx_1%7D%7B2%7D%29%2C%28%5Cfrac%7By_2%2By_1%7D%7B2%7D%29%20%5C%5C%20BD%3D%28-%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B9%7D%7B2%7D%29%20%5C%5C%20AC%3D%28-%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B9%7D%7B2%7D%29%20%5Cend%7Bgathered%7D)
Since they have the same midpoint, the answer will be;
We Prove that AC and BD have the same midpoint which we have done. we know the diagonals of a parallelogram bisect each other; therefore, they have the same midpoint and ABCD is a parallelogram.
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Answer:
only five points uneceptableeeeee