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Draw a circle. Now pick two points (A,B) on the periphery of that

  1. Aug 28, 2008 #1
    Draw a circle. Now pick two points (A,B) on the periphery of that circle, so that a line between A and B will have to pass through the center of the circle. The line AB is, in other words, the diameter of the circle.

    Now, pick a point C anywhere on the periphery of the circle. Prove that in any such triangle ABC, angle C is 90 degrees.

    --

    I have no idea. At first I was even reluctant to agree that C is always 90 degrees, but making random points on a circle seem to confirm this.

    The best I can come up with is that if I draw a normal from point C to AB, then that normal will always make 90 degree angles, so both "split-triangles" must be similar and so the original ABC must be similar as well, and contain a 90 degree angle.

    I don't think that constitutes a proof.
     
  2. jcsd
  3. Aug 28, 2008 #2

    rock.freak667

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    Re: Proof

    Put O as the centre of the circle and draw in the line OC. Noting that OC is a radius,as well as OA and OB. Put angle OAC=a and OBC=b. Now just use some simple triangle properties and you should get it.
     
  4. Aug 28, 2008 #3
    Re: Proof

    I think that since OC and AC are both R long, angle ACO must be equal to angle A.
    And for the same reason, angle BCO must equal B.

    Which means angle C should be A+B.

    A+B+C = 180
    A+B+(A+B) = 180
    2(A+B) = 180
    A+B = 90
    C = 90

    That seems to work. Thanks a lot rock.freak, I don't think I would ever have thought to split the triangle between C and O, since the new ones seem so random :)

    k
     
  5. Aug 28, 2008 #4

    HallsofIvy

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    Re: Proof

    There is a fairly well know theorem that an angle with vertex lying on a circle has angle equal to 1/2 the arc it subtends. If the two points are ends of a diameter, then the arc is subtends is 1/2 a circle, 180 degrees, so the angle is 90 degrees.
     
  6. Aug 28, 2008 #5

    tiny-tim

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    Hi kenewbie and HallsofIvy! :smile:

    Yes … the method of proof is exactly the same, plus exterior angle of triangle = sum of the two opposite angles. :wink:
     
  7. Aug 28, 2008 #6
    Re: Proof

    Ivy, ok. I've never learned that, perhaps it comes later.

    Now I know why :)

    k
     
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