Thread: 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

Originally Posted by the undertaker

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

...

Hello, the undertaker!

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

Prove that: .$\displaystyle \angle B + \angle D + \angle F = 360^o$

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.

$\displaystyle \begin{array}{ccccccc}\angle B &=& \frac{1}{2}\overline{CEA} & =&\frac{1}{2}(\overline{CE} + \overline{EA}) & [1] \\ \\ [-3mm]

\angle D &=& \frac{1}{2}\overline{EAC} &=& \frac{1}{2}\left(\overline{EA} + \overline(AC}\right) & [2] \\ \\[-3mm]

\angle F &=& \frac{1}{2}\overline{ACE} &=& \frac{1}{2}\left(\overline{AC} + \overline{CE}\right) & [3] \end{array} $



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

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

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

. . . . . . . . . . . . .$\displaystyle =\;360^o$