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Topic: A question about straight lines
Replies: 103   Last Post: Feb 8, 2014 8:28 AM

 Messages: [ Previous | Next ]
 Neighbor Posts: 57 From: It doesn't matter Registered: 1/26/14
Re: A question about straight lines
Posted: Jan 29, 2014 10:53 AM

> I hesitate to butt in, but:
> I gather we're actually discussing segments, so - if
> my memory serves me
> correctly, a segment is commonly bisected by
> constructing the
> perpendicular determined by the intersection of arcs
> the end-points of the original segment. so:
> if AB is thus bisected to AC and CB, the common end
> point of AC & CB is C.
>
> Gary Tupper
> Terrace BC

Yes, but when we want to take the two equal straight-lines (in which AB is said to have been divided through C), being C one indivisible point, how do we take them?
That's what I mean.
Let's make the issue clearer: When we want to divide something into two exactly equal parts, in the process of division we, of course, don't lose parts of the thing we are dividing, but we divide the whole in two exactly equal parts. For instance, if we have 6 candies, and we want to divide them in half, we take three on one side and three on the other, not losing candies in the process of division. Now when we divide a straight-line into two equal ones, in the process of division we lose a part of that straight-line (i.e. the point through which we divide it), that is to say, initially we have AB and finally (after the division being made) we have A(C) C (C)B, and A(C) and (C)B are said to be the two equal straight-lines in which AB has been divided, for the point C cannot be divided between them, having no parts. Let's make an example like the previous: having 5 candies that we want to divide into two exactly equal parts, we take 2 on one side and 2 on the other, and the o!
ne candy left, if it's able to be divided in half, then we will divide the 5 candies into two exactly equal parts with success. But if the candy left cannot be divided, as the remaining point C which we cannot divide, then our purpose of dividing the 5 candies into two exactly equal parts is unreachable. And the same follows for a straight-line that is divided by means of a point on itself. Therefore, the inference seems to me to be that a straight-line cannot be divided into two exactly equal parts by means of a point on it.

Date Subject Author
1/26/14 Neighbor
1/26/14 kirby urner
1/27/14 Neighbor
1/27/14 Robert Hansen
1/27/14 Bishop, Wayne
1/27/14 kirby urner
1/27/14 Neighbor
1/27/14 Robert Hansen
1/27/14 Neighbor
1/27/14 Neighbor
1/28/14 Neighbor
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Robert Hansen
1/28/14 Neighbor
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1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Neighbor
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Robert Hansen
1/28/14 Neighbor
1/28/14 Gary Tupper
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Louis Talman
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/29/14 Robert Hansen
1/29/14 Bishop, Wayne
1/29/14 GS Chandy
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1/29/14 Neighbor
1/29/14 Joe Niederberger
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1/29/14 Joe Niederberger
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1/29/14 Joe Niederberger
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1/29/14 Joe Niederberger
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1/31/14 kirby urner
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1/31/14 Neighbor
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1/31/14 Neighbor
1/31/14 Robert Hansen
1/31/14 Joe Niederberger
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1/31/14 Joe Niederberger
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1/31/14 Neighbor
1/31/14 Robert Hansen
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1/31/14 GS Chandy
1/31/14 GS Chandy
1/31/14 Joe Niederberger
2/1/14 GS Chandy
2/1/14 Robert Hansen
2/1/14 kirby urner
2/1/14 Robert Hansen
2/1/14 Joe Niederberger
2/1/14 Robert Hansen
2/1/14 Joe Niederberger
2/2/14 Robert Hansen
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2/1/14 Joe Niederberger
2/2/14 Robert Hansen
2/2/14 James Elander
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2/2/14 GS Chandy
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2/2/14 Robert Hansen
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2/8/14 GS Chandy