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A problem in geometry

  1. Sep 7, 2004 #1

    in triangle ABC a half circle is inscribed.
    the circle's diameter KL, is parallel to side AB
    CD is the height of the triangle ABC
    CD = 14 cm
    AB = 10 cm
    prove that ED equals to the circles radii
  2. jcsd
  3. Sep 7, 2004 #2
    Is there not only one radius ?
    Besides, if you don't have to calculate the radius, this is pretty straightforward.
  4. Sep 7, 2004 #3
    in the drawing there are no radii's only the diameter KL
    also the center of the circle is not given.
    im pretty sure that whats needed to be proven is that ED = KL/2
    i just don't know how...
  5. Sep 7, 2004 #4
    ED equal the circle radius because it is just the distance from the diameter to the parallel tangent to the circle. I don't need to calculate the radius here. Maybe you tricked me to get the answer...
  6. Sep 7, 2004 #5
    no i didnt trick you that might be one way to prove it but there should be a proof a bit more complex...
  7. Sep 7, 2004 #6
    Doubtlessly. Do you need to actually compute the radius ?
  8. Sep 7, 2004 #7
    no as the question states i need to prove that ED equals to the circles radii
    (in a diffrent way from the one you described)
  9. Sep 7, 2004 #8
    Ok, humanino was right though...

    since CD is the height of the triangle then CD and AB are perpendicular

    which means that angle EDB = 90degrees

    As well We know that a line tangent to a circle is perpendicular to the radius

    In the drawing FG is perpendicular to AB so FGD = 90 degrees

    as well FG and ED is perpendicular to KL since KL is parallel to AB

    DEL and GFK = 90 degrees

    so this proves that DEFG forms a rectangle with ED and FG being opposite sides thus being equal. FG equals the radius of the half circle, thus ED = radius of circle

    Attached Files:

    Last edited: Sep 7, 2004
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