Help with geometry proof

[Bobmuhthol]

I have absolutely no idea how to even start this and have been trying for days.

"The segments joining the midpoints of any quadrilateral form a parallelogram. Prove this theorem and in your proof include a restatement and a diagram."

I think it has something to do with the quadrilaterals diagonals but other than that I don't know what to do at all. I don't even have a given.

[Edaarin]

Geometry is seriously harder than econometrics.

[Latrinsorm]

Start by drawing a bizarre quadrilateral. Inscribe a diamond. I'll try to attach a gif.

[SpunGirl]

My only input is that geometry sux. I remember a test I had as a freshman, the question was this:

"Joe opens his compass to 35 degrees and makes a circle. Now he wants to make a bigger circle. What should he do?"

I answered, "open his compass wider."

I GOT IT WRONG.

-K

[Latrinsorm]

I think I'm on to something, but it doesn't quite work. New pic:

[Edited on 1-31-2005 by Latrinsorm]

[Bobmuhthol]

I'm pretty fucked. I'm probably the only one in the class that can't do it and nobody I know can.

[peam]

Before geometry, men lived for hundreds of years.

[Alarke]

What I personally would do from that point is cut the diamond into half both ways and change it into two triangles. Proof #1 = the triangle has 180 degrees (obviously). Proof #2 - send triangle is 180 degrees too. Proof #3 - something along the lines of the opposite sides of the opposite triangles creates an angle equal to the opposite side, prooving equal length in the diamonds sides as well as parallel since they have the same angle.

[Bobmuhthol]

I have no idea what that means.

[HarmNone]

It means that geometry sucks, Bob.:P

[HarmNone]

Here's a couple of sites you can have a look at to see if they give you any help, Bob:

http://www.tenet.edu/teks/math/clarifying/geometry/art.pdf
http://www.qesnrecit.qc.ca/mst/math/m536/536vCabri.pdf

[Jonty]

Originally posted by SpunGirl
My only input is that geometry sux. I remember a test I had as a freshman, the question was this:

"Joe opens his compass to 35 degrees and makes a circle. Now he wants to make a bigger circle. What should he do?"

I answered, "open his compass wider."

I GOT IT WRONG.

-K

:wtf:

What was the answer?

[HarmNone]

Yippee! Kranar's here! The answer will be forthcoming, I'm sure. ;)

[Bobmuhthol]

The first link didn't help at all.

BUT THE SECOND LINK SAVED ME AND I FUCKING LOVE YOU, HARMNONE. :heart::heart::heart::heart::heart::heart::heart:: heart::heart::heart::heart::heart::heart:

[HarmNone]

Wonderful, Bob! Glad to be of service, even if I didn't have the foggiest idea what they were talking about...:D

[Bobmuhthol]

I'm going to see if this makes sense before I write it down.

Given: ABCD with midpoints EFGH and diagonal AC
Prove: EFGH is a parallelogram

Statements:
1. Midpoints EFGH
2. EF is the midsegment of triangle ABC;
GH is the midsegment of triangle ADC
3. EF is 1/2 of AC;
GH is 1/2 of AC
4. EF parallel to AC;
GH parallel to AC
5. EF = GH
6. EF is congruent to GH
7. EF is parallel to GH
8. EFGH is a parallelogram

Reasons:
1. Given
2. Definition of midsegment
3. Midsegment theorem
4. Midsegment theorem
5. Transitivity
6. Definition of congruence
7. Two lines parallel to the same line are parallel
8. If one pair of opposite sides of a quadrilateral are both congruent and parallel, the quadrilateral is a parallelogram

[Edited on 1-31-2005 by Bobmuhthol]

[HarmNone]

Makes sense to me, Bob, considering what I saw on the site. I think you've got it solved. :)

[SpunGirl]

Originally posted by Jonty

:wtf:

What was the answer?

It was some shit about opening the compass to a degree wider than 35 degrees (I guess simply "wider" wasn't good enough) because of some theorem about the diameter being related to the degree of the compass and if the degree was bigger then the diameter would be too.

I guess the common sense answer just wasn't clusterfucked enough for them. Pythagoreus (sp?) and Euclid can rot in hell.

-K

[Artha]

I paid attention maybe 3 times in Geometry and passed with a high B/low A average. I love the greeks.

The only reason I passed Geometry in High School was because the teacher's niece loved me and begged to pass me. It was awesome since I got the lowest final in the whole class.

- Arkans

[Edaarin]

Math, my dear boy, is merely the lesbian sister of biology.

[GSTamral]

The simpler side of this is simply to say that if you are taking the midpoints, the traingular area being cut off is an isocolese (sp) triangle, because two sides are equal. Working from the 4 triangles, it must be parralel.

[Alarke]

Exactly. That site proof is what I said Bob, so it makes sense to me.

[HarmNone]

Rarely does a highschool geometry teacher want a simple explanation, even when dealing with the simplicity of an isosceles triangle. They normally want to see a more complex explanation of the student's solving process. I think Bob did an excellent job with his. :)

[Snapp]

Wow, I musta forgot a lot of stuff since high school. I remember being good at Geometry, but reading all that made my head hurt.

[Bobmuhthol]

<<The simpler side of this is simply to say that if you are taking the midpoints, the traingular area being cut off is an isocolese (sp) triangle, because two sides are equal. Working from the 4 triangles, it must be parralel.>>

I don't think you have any idea what you're talking about. Connecting the midpoints of a quadrilateral forms a parallelogram. There's no triangular area anywhere. The one diagonal separates the quadrilateral into two triangles. Neither of them are isosceles and they are not congruent unless the quadrilateral is a parallelogram.

Your misspellings:
triangular
isosceles
parallel

Please never tell me it's easy to excel in both algebra and geometry again.