imagine a uhhh a quadrilateral:

i ask you to prove the quadrilateral to be a parallelogram:

are my steps right?

1) based off of the defintion of a parallelogram: opposite pair of sides are parallel lines and diagonals are not perpendicular

you would have to prove the sides of the parallelogram to be parallel to each other (not all of them only the opposite ones) and prove that the diagonals are NOT perpendicular, meaning: the distance of the diagonals SHOULD NOT be the same and the slope of the diagonals should NOT be perpendicular....

now we write out the proof:

\(|\overline{AB}| // |\overline{CD}|\)

\(|\overline{CA}| // |\overline{DB}| \)

\(|\overline{CB}| \neq |\overline{DA}|\)

is this incorrect or correct?

and also what does the "absolute" value looking thing around the line segments mean, am i using them properly or no

thank you so much

:D

Nirvana Nov 8, 2019

#1**0 **

also how would you prove if a quadrilateral is a rhombus and a square?

im really really really confused ):

Nirvana Nov 8, 2019

#2

#4**+1 **

I know less than you lol. I am researching as I post this.

There are actually multiple ways to prove it, do you HAVE to use your theorem (1)?

Prove that both pairs of opposite sides are parallel.

Prove that both pairs of opposite sides are congruent.

Prove that one pair of opposite sides is both congruent and parallel.

Prove that the diagonals bisect each other.

Also, are you sure Diagonals can't be perpendicular? Because perpendicular diagonals form squares, and squares are parrellelograms.

CalculatorUser
Nov 8, 2019

#5**+1 **

Also, I think the Absolute value means that the length or magnitude is positive, because segments can't be negative.

CalculatorUser
Nov 8, 2019

#6**0 **

yes im sure parallelograms do not have perpendicular diagonals,,, only rectangles, some rhombi, and squares do so.

but why would you have to prove that your diagonals bisect each other? I thought that only applied to rhombi...

Nirvana
Nov 8, 2019

#10**+1 **

Is this theorem given?

"_{ opposite pair of sides are parallel lines and diagonals are not perpendicular "}

I am kind of confused, for your proof, is it based on parellelogram ABCD?

If it is based on ABCD, you have to check the corresponding segments.

Also, \(|\overline{CB}|\neq|\overline{DA}|\) is incorrect I think. Because that is saying that they are not equal in LENGTH. You want to prove that they are NOT perpendicular.

So you have to do this I think. \(|\overline{CB}|\not\perp|\overline{DA}|\)

CalculatorUser
Nov 8, 2019

#11**0 **

yea it's a parallelogram

also they aren't equal in length though,,, i thought for a parallelogram i had to prove diagonals non congruent (because it might interfere with the def. of rectangle) and opposite sides parallel (which is the same for rectangles except rectangles have congruent diagonals)

maybe i have to prove they bisect each other as well as to the other proofs though,,,

so would i use the two diagonals to find where they intersect and the midpoint is the proof?

Nirvana
Nov 8, 2019

#12**+1 **

Of course its a parrelelogram, but is it in the order ABCD?

Because your correspondences dont seem to make sense.

It is supposed to go in alphabetical order like the image above.

Look at your proof, then look at the image above, does it make sense?

CalculatorUser
Nov 8, 2019

#14**+1 **

If it goes in ABDC, then your proof should be right.

The diagonal part, I am not so sure, you might have to get someone who is experienced in Honors Geometry.

Do you have the exact problem? if you can, try to post a picture.

I might have to go soon, sorry, I hope CPhill or Melody or someone educated can help!

CalculatorUser
Nov 8, 2019