# Thread: circle geometry

1. ## 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

2. 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

...

3. 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]

\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]:

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

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

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