Hello, Natasha!

Some basic geometry (plus a little algebra) is required.

A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

The vertices of a cyclic hexagon are labelled in order A to F.

Prove that the sum of the interior angles at A, C and E

is equal to the sum of the interior angles at B, D and F. Code:

A
* o *
* o o *
F o o B
*o o*
o o
* o o *
* o * o *
* o o *
o o
*o o*
E o o C
* o o *
* o *
D

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

. . $\displaystyle \angle A \;=\;\frac{1}{2}\left(\widehat{BC} + \widehat{CD} + \widehat{DE} + \widehat{EF}\right)$

. . $\displaystyle \angle C \;=\;\frac{1}{2}\left(\widehat{DE} + \widehat{EF} + \widehat{FA} + AB\right)$

. . $\displaystyle \angle E \;=\;\frac{1}{2}\left(\widehat{FA} + \widehat{AB} + \widehat{BC} + \widehat{CD}\right)$

Then: .$\displaystyle \angle A + \angle C + \angle E \;=\;\frac{1}{2}\left(2\!\cdot\!\widehat{AB} + 2\!\cdot\!\widehat{BC} + 2\!\cdot\!\widehat{CD} + 2\!\cdot\!\widehat{DE} + 2\!\cdot\!\widehat{EF} + 2\!\cdot\!\widehat{FA}\right)$

Hence: .$\displaystyle \angle A + \angle C + \angle E \;=\;AB + BC + CD + DE + EF + FA \;=\;360^o$

In a similar fashion, it can be shown that: .$\displaystyle \angle B + \angle D + \angle F \;=\;360^o$