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Math Help - Euclidean Geometry

  1. #1
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    Euclidean Geometry

    Euclidean Geometry-geometry-2.1-13-17.pdfThe attachment consists of 2 exercises from my last week's homework that I attempted to prove using numerous strategies, but never found a good one.

    I searched for a while for the characteristics of a cyclic quadrilateral to use for the set up of the first proof, but never got even that far, and thus didn't get very far at all (I know...No S*^%!). As for the second exercise, no axiom, theorem, definition, or combination seemed very strong to prove the part i proposition. That left part iii pretty hopeless.

    Thanks!
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  2. #2
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    According to PlanetMath, a necessary and sufficient condition for a quadrilateral to be cyclic, is that the sum of a pair of opposite angles be equal to 180 degrees.

    Concerning part 1, let F be a point on m left of C. I believe there is a theorem that angle FCA = angle ABC (if you need help finding it, let us know). We have angle ADE = 180 - CDE = 180 - FCA. Thus, ADE + ABE = 180. Similarly, it's easy to show that DAB + DEB = 180.
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  3. #3
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    Curious about the theorem

    Thank you very much for the help emakarov! Please forgive me if I'm being dense, but the claim that angle FCA = angle ABC is difficult for me to accept given that F is an arbitrary point on m, only constrained such that it lies to the left of C. Because angle ABC is fixed once constructed, my wee brain cannot find a satisfactory way to believe the proposition.

    I searched a bit for the theorem you mentioned, but didn't get far because I really don't know what I might search for, i.e. what to guess that the theorem might be called.

    Notwithstanding all that, thanks for your help.
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  4. #4
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    Put F on m so that A~\&~F are on the same side of \overleftrightarrow {BC}

    Now angles \angle ABC\;\& \,\angle FCA are congruent, they intercept the same arc.

    Angles \angle CDE\;\& \,\angle FCA are alternant interiors so are congruent.

    Angles  \angle CDE\;\& \,\angle EDA are supplementary.

    Therefore angles  \angle ABE\;\& \,\angle EDA are supplementary.

    That proves the quadrilateral is cyclic.
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  5. #5
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    the claim that angle FCA = angle ABC is difficult for me to accept given that F is an arbitrary point on m
    Since F is on m and left of C, moving it does not change the angle FCA!

    As Plato said, \angle ABC and \angle FCA are equal because they are half the arc AC. I found this explanation on cut-the-knot.org. You can also search for "inscribed angle".
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  6. #6
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    Thanks! Your responses led me to finally getting the idea through my rather thick skull.

    I'm about to post another question that I find really interesting. After a good bit of consideration, I haven't come up with a solution for. Hope you check it out.

    Thanks again!
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