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Not exacly the place for it... Math help
Posted: Sun Jun 10, 2007 11:59 pm
by anthony88guy
Instead of finding a math forum, registering and that whole tid bit, I figured ill try my query here first. Even if someone could just tell me whats this called besides transversal lines, I would have some luck finding something helpful on the web.
So here we go a geometry question:

(not to scale)
As stated in the image AC and DB are parallel lines.
AB = 14
AC = 3
DC = 4
Now we have to find AE.
I remember having to set up a proportion? Something along the lines of:
But we don't have enough information to do so.
Any help would be greatly appreciated.
Thanks.
Posted: Mon Jun 11, 2007 12:24 am
by feyd
Looks like an application of the Law of Sines and Law of Cosines.
Posted: Mon Jun 11, 2007 8:49 am
by Oren
ACE =~ BDE, thus:
AC/DB = AE/BE
AC = 3
DB = 4
=>
3/4 = AE/BE
Let's write AE = x
AE = x; AB = 14 => BE = AB - AE = 14 - x
=>
3/4 = x/(14 - x)
=>
AE = x = 6

Posted: Mon Jun 11, 2007 9:04 am
by superdezign
Oren wrote:AE = x; AB = 14 => BE = AB - AE = 14 - x
^_^ I need to get back into math class. That thought didn't even come to mind.
Posted: Mon Jun 11, 2007 9:19 am
by Oren
Actually it should be like this (but who cares):
1. Denote: AE = x
2. AB = 14
3. BE = AB - AE
=>
BE = 14 - x
P.S I didn't give full proof as this is not a math test (e.g I didn't prove that ACE =~ BDE since it's too easy).
Posted: Mon Jun 11, 2007 10:36 am
by anthony88guy
Thank you Oren.
I knew it was a proportion problem just couldn't figure out how to set it up.
Posted: Mon Jun 11, 2007 11:11 am
by RobertGonzalez
This also fits into the category of similarity. The fact that AC||DB means that triangle ACE is similar to triangle BDE (provided segment AB and segment CD are linear). Given the similarity, you can set up proportionate equality to solve, as Oren did.
Posted: Mon Jun 11, 2007 1:48 pm
by Oren
Everah wrote:This also fits into the category of similarity. The fact that AC||DB means that triangle ACE is similar to triangle BDE (provided segment AB and segment CD are linear). Given the similarity, you can set up proportionate equality to solve, as Oren did.
Everah, that's exactly what I used. "ACE =~ BDE" means "triangle ACE is similar to triangle BDE"
What do you mean by the way when you say "provided segment AB and segment CD are linear"? AC||DB that's enough.
Posted: Mon Jun 11, 2007 3:08 pm
by RobertGonzalez
I know that is what you meant. I just wasn't sure the OP knew that =~ means similar.
And I suppose by linear I was just complicating things. I was thinking that if the intersecting segments were curves, then it would throw similarity out the window, but I think I am just being weird (and showing my age a little. I haven't been deeply involved in Geometry for about 18 years now).
Posted: Tue Jun 12, 2007 7:16 am
by Oren
Everah wrote:I haven't been deeply involved in Geometry for about 18 years now.
Wow, and yet you remembered? That's great man... seriously. I can be a master in many different things, but if I don't use them for few months then I totally forget them. I'm really amazed that you remembered that after nearly 18 years

Posted: Tue Jun 12, 2007 10:09 am
by RobertGonzalez
Math was the one thing that I really enjoyed throughout my education. I took math all the way through linear equations and differential calculus. Of course, that was 13 years ago (when I completed my math learning). Geometry was my sophomore year in high school.
PS I just realized how old that makes me sound.

Posted: Tue Jun 12, 2007 10:35 am
by patrikG
Are you guys helping some guy with his homework again?

Posted: Tue Jun 12, 2007 11:24 am
by RobertGonzalez
Hell no. I purposely muddled things up to make him go back to his text book and learn it the right way.
