On the same straight line there cannot be constructed two....

[astrololo]

So, according to this figure :http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII23.html We cannot have similar segments of circles and unequal ones be built on the same side of the same straight line.

My question is : Can we build similar segments of circles but unequal ones ? (It seems to imply it)

The definition of similar segments of circles is : "Similar segments of circles are those which admit equal angles, or in which the angles equal one another."

Here is the link for it : http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/defIII11.html

Thank you!

 

[FactChecker]

Are you asking if it is possible when you omit "same side"? I would say yes, because any diagonal has two similar, unequal segments on opposite sides.

 

[Merlin3189]

... I would say yes, because any diagonal has two similar, unequal segments on opposite sides.

I'm just wondering what you mean by "diagonal" here? Is it the same as "chord" ?
If you are referring to the segments on the opposite side of a chord, they do not seem to be similar as the subtended angles are supplementary rather than equal.

 

[FactChecker]

I'm just wondering what you mean by "diagonal" here?

Oh, I meant diameter. Thanks for the correction.

 

[Merlin3189]

Well, if it's a diameter, then the two parts would be identical, so similar but not unequal. Putting them on opposite sides doesn't help, they are still equal.

Going back to the OP I share his bewilderment: I can't see why which side of the line they are makes any difference. Putting unequal segments on opposite sides would never make them similar. As far as I can see, the only reason for drawing them on the same side is to make the diagram for his proof work.
But to me it is a strange proof, using a non-obvious fact (angles in a segment are equal) to prove an obvious one!

As far as the question, "Can we build similar segments of circles but unequal ones ?" goes, if the line AB means the line starting at A and ending at B, and this has to be the chord of the circle, then no.
To me, "the line AB" has always meant "the line which passes through A and B, extended infinitely in both directions". In that case we could have any number of similar segments (enlargements of each other) which need not be equal sitting on that line on either side.
Just as similar triangles (or any other shapes) with equal bases are equal, then similar segments with equal chords are equal. So you could not have unequal similar shapes with the same base, whichever side of a line they were on.

 

[FactChecker]

Well, if it's a diameter, then the two parts would be identical, so similar but not unequal.

Are they talking about equal length or equal? The two are similar and equal length, but not equal (they are not the same segment). I think that is the only interpretation of the proposition that would make it true.

 

[astrololo]

By the way, the base of the segment doesn't need to be equal. My question would be, can we have two unequal segments of circles that are similar, but they don't need to have equal bases.