Hello, u2_wa!
Prove that sum of opposite angles in any cyclic quadrilateral is 180°? Code:
A
* o *
* o o *
* o o *
* o o B
o o
* o o *
* o o *
* o o *
o o
D o o *
* o o *
* o *
* o *
C
An inscribed angle is measured by one-half its intercepted arc.
We have inscribed quadrilateral $\displaystyle ABCD.$
. . $\displaystyle \begin{array}{cccc}\angle A &=&\frac{1}{2}\,\text{arc}\,(BCD) & {\color{blue}[1]}\\ \\[-3mm] \angle C &=& \frac{1}{2}\,\text{arc}\,(DAB) & {\color{blue}[2]} \end{array}$
Add [1] and [2]:
. . $\displaystyle \angle A + \angle C \;=\;\frac{1}{2}\,\text{arc}\,(BCD) + \frac{1}{2}\,\text{arc}\,(DAB) $
. . . . . . . . .$\displaystyle = \;\frac{1}{2}\,\underbrace{\bigg[\text{arc}\,(BCD) + \text{arc}\,(DAB)\bigg]}_{\text{This is }360^o} $
. . . . . . . . .$\displaystyle = \;\frac{1}{2}\,(360^o)$
. . . . . . . . .$\displaystyle =\;180^o$
In a similar manner, we can prove that: .$\displaystyle \angle B + \angle D \:=\:180^o$