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Math Help - circle geometry

  1. #1
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    circle geometry

    points A, B, C, D, E and F lie on the circumference of a circle.

    Prove that angle ABC + angle CDE + angle EFA = 360 degrees
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  2. #2
    MHF Contributor
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    Quote Originally Posted by the undertaker View Post
    points A, B, C, D, E and F lie on the circumference of a circle.

    Prove that angle ABC + angle CDE + angle EFA = 360 degrees
    Draw a circle with all the points on the circumference and the centre labelled. Also, connect each points to the centre.

    There are 6 different isosceles triangles with different base angles, call them angle 1, angle 2, angle 3, angle 4, angle 5, angle 6.

    For each triangle, the remaining angle is 180-2(base angle)

    Ex: For triangle OBC, remaining angle is 180-2(angle 1)

    The sum of all remaining angles is 360.

    180-2(angle 1)+180-2(angle 2)+...+180-2(angle 6)=360

    Simplify this, angle 1 + angle 2 +...+ angle 6=360

    angle ABC = angle 2 + angle 1

    angle CDE = angle 6 + angle 5

    angle EFA = angle 3 + angle 4

    ...
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  3. #3
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    Hello, the undertaker!

    Points A, B, C, D, E, F lie on the circumference of a circle.

    Prove that: . \angle B + \angle D + \angle F = 360^o
    Code:
                    A
                  * o *
              *  o     o  *
            * o           o *
         F o                 o B
           o                 o
          *o                 o*
          *o                 o*
          *o                 o*
           o                 o
         E o                 o C
            * o           o *
              *  o     o  *
                  * o *
                    D

    An inscribed angle is measured by one-half its intercepted arc.

    \begin{array}{ccccccc}\angle B &=& \frac{1}{2}\overline{CEA} & =&\frac{1}{2}(\overline{CE} + \overline{EA}) & [1] \\ \\ [-3mm]<br /> <br />
\angle D &=& \frac{1}{2}\overline{EAC} &=& \frac{1}{2}\left(\overline{EA} + \overline(AC}\right) & [2] \\ \\[-3mm]<br /> <br />
\angle F &=& \frac{1}{2}\overline{ACE} &=& \frac{1}{2}\left(\overline{AC} + \overline{CE}\right) & [3] \end{array}



    Add [1], [2] and [3]:

    \angle B + \angle D + \angle F \;=\;\frac{1}{2}\left(2\!\cdot\!\overline{AC} + 2\!\cdot\!\overline{CE} + 2\!\cdot\!\overline{EA}\right)

    . . . . . . . . . . . . . =\; \overline{AC} + \overline{CE} + \overline{EA}

    . . . . . . . . . . . . . =\;360^o
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