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Thread: Modern Geometry: Triangle, perpendicular bisectors, circumcircle, circumcenter

  1. #1
    Member ilikedmath's Avatar
    Sep 2008

    Question Modern Geometry: Triangle, perpendicular bisectors, circumcircle, circumcenter

    Write out a complete proof of the following:

    THEOREM: The perpendicular bisectors of the sides of a triangle, if 2 of them meet, are concurrent at a point that is the center of a circle (circumcircle) passing through the vertices. (The point of concurrency is called the circumcenter of the triangle.)
    NOTE: The conditional is needed here because in absolute geometry it is possible for 2 such perpendicular bisectors to be non-intersecting lines.

    My work:
    A picture I found to illustrate the theorem:

    (Side note: I don't understand what they mean by "the conditional is needed here...." Let me add that my professor is very confusing, and I am quite lost in this class. I read the book, but it doesn't help.

    I don't even know where to begin. So what exactly am I being asked to prove? Is it to prove that the point O is the center of a circle? I know that
    $\displaystyle \overline{AR}$ = $\displaystyle \overline{RC}$, $\displaystyle \overline{CP}$ = $\displaystyle \overline{PB}$, and $\displaystyle \overline{BQ}$ = $\displaystyle \overline{QA}$ by definition of perpendicular bisector.

    But then, I can't quite see how knowing those will help me. As I stated in my previous post, I am unsure as to exactly what tools we can/can't use to proving statements because we are approaching geometry in a very different method than how we learned in high school. We are to basically "unlearn" what we were taught in high school, and build our proofing tools. I am just utterly lost.

    Any help at all is greatly appreciated. Thank you very much for your time.

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  2. #2
    Jan 2009
    This is a basic theorem and there should be formal a proof somehere in internet. Let's try something a little clumsy but it MAY work:

    1) Extend OP in both directions. Let M and N be the points where extended OP intersects the circle.

    2) We know that:
    |MP|*|PN|= |BP|*|PC| (This can be shown by similarity of triangles)
    => |MP|*|PN|= |PC|^2 (Can you see the Euclidean identity in MCN triangle?)
    => angle(MCN) = 90 degrees.
    => MN is the diameter

    3) Repeat the same steps for OQ, which will reveal that O is the center of THE circle.

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