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