Hi guys can i please get help with these questions urgently as soon as possible
Thanks
Before I begin, I would like to disclaim that there is a lack of mathematical rigor associated with this proof. It was around 12:00AM when I wrote this. I recommend that you are more rigorous than I am when constructing your own proof.
Q7. Let there be a quadrilateral ABCD
Let
, the midpoint of \vec{BC} = \frac{1}{2}b (denoted by DX), the midpoint of \vec{CD} = \frac{1}{2}c (denoted by BY), the midpoint of \vec{DA} = \frac{1}{2}d (denoted by CN).<br />)
It is known, as ABCD is a quadrilateral, that:
Therefore,
)
Generating other items of information for our midpoints:
$$ \vec{MY} = -\frac{1}{2}a- \frac{1}{2}d $$
$$ \vec{YN} =- \frac{1}{2}d -\frac{1}{2}c $$
$$ \vec{XN} = \frac{1}{2}b+ \frac{1}{2}c $$
$$ \vec{MX} = \frac{1}{2}a + \frac{1}{2}b $$
If MYNX is a parallelogram, then:
and
Thus,
(1)
[tex ]-\frac {1}{2}a - \frac{1}{2}d = \frac{1}{2}b + \frac{1}{2}c[/tex]
(2)
We are already aware that (2) is true due to the following:
<br />)
We can also demonstrate that (1) is true from basic tranposition (giving the same result as (2)).
Therefore, the midpoints, when joined, form a parallelogram.
You can use this result to find the solutions to your other questions.
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