Kiselev Planimetry And Stereometry

[Zee Palmer]

Iam from the Czech Republic and thus my Math education was nearly completely proof-less. Thus I have started reading How to Prove It by Velleman for general proofs and Kiselev's Geometry to brush up my planimetry / stereometry (which I lack intention for and generally I'm pretty bad at, even tough the rest of high school math is easy for me and I'm want to study Math at a university) and to learn how to do geometrical proofs.

I'm stuck on exercise 55. I have tried to work with the definitions from the preceding chapter but to no avail. It maybe because I have nearly zero experience with geometrical proofs, so any tips and tricks would be highly appreciated.

Second, you should stop using the expression "do proofs". There is no such an activity in mathematics as "doing proofs": we are concerned not with how to derive some things from other things, but what is true in the world around us, and what is false, and why.

Then I would try to construct a counter-example for quadrilaterals (even though the problem claims they don't exist), for - who knows maybe the book is wrong? After some tries, if you don't succeed, maybe you'll get some idea why. Presenting such an idea in a form neat enough to guarantee that there are no

loopholes, would constitute a solution to the problem.

1. for any vertex in any polygon, if the angle between a diagonal and a side is greater than the angle between the two sides

then the diagonal lies in the exterior of the area bounded by the two sides

2. for any vertex in any polygon, if the angle between a diagonal and a side is less than the angle between the two sides

then the diagonal lies within the interior of the area bounded by the two sides

Also, note that your sequence of statements is not really an "argument": you make claims, but don't even try to explain why they are correct

[In fact some of them are incorrect as stated, e.g. 2].

Construct the four points in the plane by placing 3 in a triangular formation with the fourth in the center of the others. 3 quadrilaterals can be constructed by alternating from which two of the outer 3 points the sides of the fourth(center) point connect.

A diagonal emanating from vertex $A$ originates interior or exterior to the quadrilateral. For it to lie both exterior and interior it would need to cross through(intersect) a side of the quadrilateral. But it has been shown that diagonal $\overlineAD$ cannot intersect any side of the quadrilateral. Thus a diagonal of a quadrilateral lies exclusively interior or exterior to the quadrilateral.

A convex broken line cannot self-intersect as this would violate the definition of being convex, the broken line lying entirely on one side of any of its continued segments. Therefore the closed convex broken line satisfies the definition of a polygon boundary.

$\S 33$ defines an isoceles triangle as having 2 congruent sides. It does not strictly state "only" 2 congruent sides. An equilateral being defined as having 3 congruent sides can be considered isoceles.

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