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Geometry: Similarity and Power of a Point 
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#1
Nov2003, 09:20 PM

P: n/a

Hello,
I am not sure how to interpret this question: So what do I have then? Lines AB and CD which intersect at P where P is not between A and B but P is between C and D? (Why don't they just draw a diagram? @#$!@#$!@#$%!#$!#$!!!!). In fact, is it even possible to have a cyclic quad. if P is between CD and not between AB? I can only see a cyclic quad. happening if P is not between both AB and CD. [?] Opinions on the question would be appreciated. Thankyou. 


#2
Nov2003, 11:07 PM

P: 1,046

My opinion is there's something wrong with the question.
If P is not between A and B, then isn't P at either A or B? And if so, don't you end up with a triangle rather than a quadrilateral? 


#3
Nov2003, 11:49 PM

P: n/a

The way I look at it is if the point of intersect is not between AB and not between CD then a simple convex cyclic quad. can be formed.
If the point of intersection is not between AB but the point of intersection is between CD then the resulting quadrilateral cannot be cyclic. Since the question states if a condition is met a cyclic quad. results, this leads me to believe that point P must not be interior to AB or CD. But if that is the case, why did the question not specify the condition for both segments, AB and CD? 


#4
Nov2103, 06:19 AM

Emeritus
Sci Advisor
PF Gold
P: 16,091

Geometry: Similarity and Power of a Point
However, you state that (XY) is the directed distance from X to Y. You can prove something about P's location WRT C and D using the directions of these segments. For instance, consider the trapezoid ABCD whose two bases are BC and DA. Draw the lines AB and CD and see where they intersect. 


#5
Nov2103, 09:04 AM

P: n/a

I see. I think I can see how a cyclic quad. can result from the above situtation then. The key is directed distances. But I am not sure if I am doing it correctly though.
Do I start with a point P and draw lengths PA, PB, PC, and PD from it? I don't remember how to use directed distances. 


#6
Nov2103, 11:41 AM

Math
Emeritus
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Thanks
PF Gold
P: 39,682

" I don't remember how to use directed distances."
Then you might want to start by looking up the definition of "directed distance". 


#7
Nov2403, 10:06 PM

P: n/a

Yes, I did. The thing is I haven't done geometry for over fifteen years. The prereq. for this course was just a first year math course  didn't matter what course as long as it was first year. This course started right in the middle text, chapter three. Directed distances were located in chapter one. So the class missed some basic geometry def. and theory (one of those things being directed distances).
Anyway, this assignment has already been submitted. And I wasn't able to do the question. But I still would like to know how to set up the question. I can do it once I get started, but I don't know how to set it up. I have looked at the def. of directed distances, but I didn' help me regarding this question. Any explanations regarding the set up with this question would be appreciated. Thankyou. 


#8
Nov2403, 10:12 PM

Emeritus
Sci Advisor
PF Gold
P: 16,091

Well, what is the definition of directed distance?
From it, you should be able to prove this lemma: If A, B, and C are collinear, then (BA)(BC) is negative if and only if B is between A and C. 


#9
Nov2403, 10:36 PM

P: n/a

Def. given me:
Let points A and B be points on the line. The directed distance AB id defined as follows: the directed distance AB is equal to AB if A < B 0 if A = B AB if B < A I will work on the lemma next. bbl. 


#10
Nov2403, 11:55 PM

P: n/a

I am not very good at proofs. But I will give this a shot.
If (BA)(BC) is negative then either: 1. (BA) is negative and (BC) is positive or 2. (BA) is positive and (BC) is negative. Assume BA is negative, then for the directed distance BA, BA if A < B. Assume BC is positive, then for the directed distance BC, BC if B < C. Therefore A < B < C. Conversely, Assume BA is positive, then for the directed distance BA, BA if B < A. Assume BC is negative, then for the directed distance BC, BC if C < B. Therefor C < B < A. In both cases, B is between A and C. That would be be my substandard proof. okokokok. I think I see where you are leading me. If I know this then I know the orientation of (PA), (PB), (PC), and (PD). In this case would A be between P and B and would C be between P and D then? I would then have two lines that intersect at P. The points A,B,C,D would be points on a circle. P is a point outside of the circle. A,B,C,D would also be the vertices of a cyclic quad. Is this a correct assessment? 


#11
Nov2503, 06:30 AM

Emeritus
Sci Advisor
PF Gold
P: 16,091

You'll have to apply one of those fun betweenness axioms if you need to pin P's location down more precisely. (In this case, if A is between P and B, then D will be between P and C) 


#12
Nov2503, 12:15 PM

P: n/a

Sorry. I wasn't thinking properly last night.



#14
Nov2503, 05:37 PM

P: n/a

Thanks for your help Hurkyl.



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