Euclidean Geometry - I need help!!!

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Simontetreault

New member
This is my first fricking class of the YEAR and the guy says \'Have you done high school geometry?\' I say yes... he says that\'s all I need.

He gave us 30 basic theromes to use to proove this, in our own words. SSA, ASA, pythagorean theorems and so on.

Does anyone know how to even begin this? I couldn\'t even draw these stupid circles by myself!!! Gah if you know how to do this or can give me a push in the right direction please let me know.




1. If CA is perpendicular to AO, DB is perpendicular to BO, and we have AC = BD
and AO = BO, labeled so that triangles OAC and OBD are directly congruent,
prove that AB bisects CD.
 
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elouchard

Guest
Geometry problems are logic problems. start by drawing the triangles and things will become clearer. Whenever a problem has various equalities between points and distances it is better to draw the scene if you are a visual thinker.
 
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Simontetreault

New member
Well

Well I believe that this is needed to make the proof. you need to have more angles, etc to make sure you can proove it. If you put them like you said it will be hard, if not impossible, to prove.
 
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elouchard

Guest
If the triangles are set up so there is a straight line from A to B, passing through O, you will have clearly bisected CD through O. Now maybe you have to rotate them and are not allowed to jump over O, but you can then see the angles are equivalent in the new triangles you are making when rotating. So if the angles are the same, the lengths are the same of the sides.
 
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Simontetreault

New member
Hmm

here is the problem, if you do it as you tell me then I don\'t think it would work because it has to BISECT AB and CD. Which means that all those lines have to be equal, which they wouldn\'t be since CO would be longer then AO and DO longer then BO.

In order for them to bisect (equal lenghts) I need to move it like that a little.

Anyway this is what I think... lol I am 100% sure our prof did not give us something so simple :)
 
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elouchard

Guest
Here is the picture. AB has to bisect CD, which it clearly does, as by defintion, the two sides of a bisected line are equal. Now if you rotate it from there you can use all sorts of angular theorums to show equivalency and solve the proof.

geom.jpg
 
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Ritual

New member
@Elouchard
But then you have not proven the statement in a general way, just the special case that you illustrated. There is nothing in the problem description that says that AO and BO (and CO and DO) are along the same line. Actually, the statement becomes trivial in that special case since the problem description already states that AO = BO and CO = DO which automatically makes the statement true.

@Simon
I\'m afraid I can\'t help you without having the theorems you can use. It\'s 10 years since I studied Euclidian geometry so I don\'t have those fresh in my mind... :innocent:
 
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lono

New member
Originally posted by darkartminiatures
This thread is making me feel really stupid :eek: Best of luck BTW
Click to expand...
No, no, you see the majority of us don\'t get it, so it\'s not us that are stupid, it\'s geometry!

Geometry is stupid!

I guess that means that you should drop it Simontetreault. Study something useful like Latin instead!

And I don\'t know if this will be of use, but I seem to remember that a square of hippopotamuses is equal to some of the squares of another two Cydes... I don\'t know who Cyde is though. Perhaps a new Daikinee Elf from Rackham.
 
airhead

airhead

Coffin Dodger / Keymaster
Originally posted by Simontetreault
1. If CA is perpendicular to AO, DB is perpendicular to BO, and we have AC = BD and AO = BO, labeled so that triangles OAC and OBD are directly congruent, prove that AB bisects CD.
Click to expand...
CA perpendicular to AO = angle
BD perpendicular to BO = anlge (also)

AC=BD = side
AO = BO = side

ACO has SAS given.
BDO has SAS given.
also given are the angles are the equal and the sides are the equal.
:: the triangles must be equal.

You need to take a quick course on Analytic Geometry to understand proofs.

Surveying is applied geometry.
Almost everything involved in design uses geometry to some degree.
 
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elouchard

Guest
Originally posted by Ritual
@Elouchard
But then you have not proven the statement in a general way, just the special case that you illustrated. There is nothing in the problem description that says that AO and BO (and CO and DO) are along the same line. Actually, the statement becomes trivial in that special case since the problem description already states that AO = BO and CO = DO which automatically makes the statement true.
I realize that but I did not want to answer the question for him. I just wanted him to see the answer visually before trying to piece it together from proofs. Maybe that is not how he thinks things through but it helps me to draw things simply, then do the rotations and angular translations.
Click to expand...
Click to expand...
 
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Highbulp Billy

New member
Wow, who\'d have thought you\'d have found geeks on a mini-painting site? ;)

Seriously though, I used to think I knew about maths but now I thank God that I never studied it beyond GCSE (though I did the advanced exam and passed). I do know that the square of the hippotomus is equal to the square of its bum on both sides or something similar. Still, pretty pictures, you could use them to make a windmill :p
 
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matty1001

New member
Yep COD and AOB are straight lines therfore the four lines (AO BO CO and DO are all equal, and hevae equal angles at ACO BDO DBO and CAO, kinda like that, hard to explain, iv not done any maths since after year 11 at school! But that should be the general jist of things.

I think you made it to complicated for your self by saying that the tutor didnt give you something so simple, a rod in your own back me thinks!
 
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Ritual

New member
Originally posted by matty1001
Yep COD and AOB are straight lines
Click to expand...
Not according to the facts in the problem description. They CAN be straight lines, but they can also be like in the picture Simon posted. So, to prove the statement you can\'t assume they are straight (in which case there\'s nothing to prove)!
 
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matty1001

New member
Ah bugger it then, i was only trying to sound clever :rolleyes: honest!

Thats just what i came up with, easier to just make things simpler if you ask me. lol
 
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Ritual

New member
@matty
Simplest thing is to not bother with geometry! lol

@Simon
I\'ve solved this (yeah, boring day at work... :rolleyes: ) so if you want help PM me!

EDIT: No, sorry... :innocent: I had made a mistake... I didn\'t solve it!
 
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