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Hi all. Does somebody know any source from where we can get the solutions to exercises of this book? Or at least, we could get proper hints or answers to selected odd numbered problems? I've started reading this book and exercises look me really technical. So I thought to solve all of these problems. I've started from Introductory chapter and its exercises.

People with similar interest are warmly welcome to join me here so that we can share ideas or match our answers, that's a recommendation from the author of this book. Waiting for your replies.

Thank you,

Raja.

*Last edited by Raj.01 (2015-08-07 16:38:54)*

Life is the process of narrowing down the probabilities.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I have never seen a solutions manual for that book. Post the problems that are troubling you.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Hi Bobbym, right now I don't have any issue with some problem and I'm writing the solutions smoothly but it's the way to match my answers with someone else. I've setup a free wordpress blog on this that's dedicated mainly to this book. I don't know if I'm allowed to share it here or not, but I've it on my public profile here.

If someone is interested, I can also copy and paste all those answers here but obviously it'll take a lot of time.

Life is the process of narrowing down the probabilities.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Okay, but if you get stuck post.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Why did you pick Kiselev's book?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Actually, truly speaking, my basics were very week when it comes to geometry, so I needed a book to study it in my spare time, so I searched on web and I heard about this book a lot. One interesting thing was that, it was being used as a text book for several decades in various countries. So I picked it up and started studying it. There's not any further reason why I picked it up.

In combination, I've two more book as well, that I keep parallel while studying geometry. But if you know some better book(s), please share with me, I feel much pleasure when I get some interesting books.

Life is the process of narrowing down the probabilities.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

If it is helping you then stick with it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Hi all, I'm stuck with this problem "Can sums (differences) of respectively congruent line segments, or arcs, be non congruent? Can sums (differences) of respectively non congruent segments, or arcs be congruent?"

I can't understand the statement. For example, "the sums of two congruent line segments are non congruent (to what?)

Please help me understand the statement. I'm confident I can solve it myself. If you've ideas, please share.

Life is the process of narrowing down the probabilities.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

hi Raj.01

Hhmmm. It does lose something in the translation. I'm puzzled too. Does it mean this:

AB is congruent to DC. BF is congruent to CG.

The second lines are added to the first. Is AF congruent to DG?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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hi Bob, I think so it is something like that what you sketched. While I read the definition of angle from this book's first page, I felt the translator's English wasn't so good. But I'm at the start of this book. if you could recommend some better books, i'll appreciate you.

Life is the process of narrowing down the probabilities.

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Hello Bob, I'm stuck with another problem. I put a lot of time solving it but couldn't do that. Please help me:

Question 157 of Kiselev's Book 1, Compute angles of a triangle which is divided by one of its bisectors into two isosceles triangles. Find all solutions.

Life is the process of narrowing down the probabilities.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

hi Raj.01

I haven't got that book. I'm having trouble getting a diagram. Please could you post one, or failing that, describe it using A, B, C etc.

Thanks,

LATER EDIT. Actually, I think I've narrowed it down to one case, but I'd still like a diagram to help confirm it.

Bob

*Last edited by bob bundy (2015-08-23 21:32:49)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

Post complete.

This is a general diagram. Lengths are not meant to be accurate. I have, however, bisected ABC accurately.

I have marked the two bisected halves with an x.

When I first read the problem, I was not sure which sides were equal in the isosceles triangles. I'm assuming that ABD and DBC are the isosceles triangles, but, for each, which are the two equal sides.

After a moment I realised that some possibilities are obviously impossible, so I decided to try all possibilities and find out which are impossible and which lead to answers. Here is my analysis:

case 1

AB = DB and DB = CB.

=> BAD = BDA = BDC = BCD = 90 - x/2

So at D, 90 - x/2 + 90 - x/2 = 180 => x = 0

So this case is not possible.

case 2

AB = DB and BD = DC

So DCB = x and so BDA = 2x.

Also BAD = 90 - x/2, so in triangle ABD

x + 90 - x/2 + 2x = 180 => 5x/2 = 90 => 5x = 180 => x = 36.

case 3

AB = BD and BC = DC

=> BDC = x and BDA = 90 - x/2

=> x + 90 - x/2 = 180 => x/2 = 90 => x = 180

So this case is not possible.

case 4

AB = AD and BC = DC

=> ADB = BDC = x => x = 90.

This case is not possible.

case 5

AB = AD and BD = DC

=> ADB = ACB = x => 2x = x => x = 0

This case is not possible.

I think that concludes all possible cases. Unless you can spot any others

Whoops. Found one myself.

case 6

AD = BD and BD = DC

So BAD = BCD = x and ADB = BDC = 2x => 4x = 180 => x = 45

Bob

*Last edited by bob bundy (2015-08-23 23:45:09)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Thank you Bob. But the question says, find all solutions. What does it mean? It means that the solution isn't unique?

Also, please tell me a name for a book/eBook that can give me a detailed oriented topics regarding plane figures only e.g. Polygons, their types, properties, theorems regarding them etc.

Life is the process of narrowing down the probabilities.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

hi Raj.01

I found two cases, x=36 and x=45. I think that's all.

At school I was taught geometry from a variety of books. I don't have a preferred one now. You could always refer to the master:

http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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hi Bob. Thank you for the book.

Can you answer me one question please? I know that a regular polygon is convex. But if the converse of this true? That is, a convex polygon is regular (in general)?

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,400

No. E.g.. any isosceles triangle.

Regular means all internal angles and all sides are equal.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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