1. sum of opposite angles in any cyclic quadrilateral is 180

How to prove that sum of opposite angles in any cyclic quadrilateral is 180? With logical reasoning.

2. 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 $\displaystyle ABCD.$

. . $\displaystyle \begin{array}{cccc}\angle A &=&\frac{1}{2}\,\text{arc}\,(BCD) & {\color{blue}[1]}\\ \\[-3mm] \angle C &=& \frac{1}{2}\,\text{arc}\,(DAB) & {\color{blue}[2]} \end{array}$

. . $\displaystyle \angle A + \angle C \;=\;\frac{1}{2}\,\text{arc}\,(BCD) + \frac{1}{2}\,\text{arc}\,(DAB)$

. . . . . . . . .$\displaystyle = \;\frac{1}{2}\,\underbrace{\bigg[\text{arc}\,(BCD) + \text{arc}\,(DAB)\bigg]}_{\text{This is }360^o}$

. . . . . . . . .$\displaystyle = \;\frac{1}{2}\,(360^o)$

. . . . . . . . .$\displaystyle =\;180^o$

In a similar manner, we can prove that: .$\displaystyle \angle B + \angle D \:=\:180^o$

3. Sir, kindly tell me what do you mean by 0.5 arc (BCD)?

4. Originally Posted by u2_wa
How to prove that sum of opposite angles in any cyclic quadrilateral is 180? With logical reasoning.
Hello..

Look at the diagram. Let the centre is at $\displaystyle O$

$\displaystyle \angle ABC=\frac 1{2}\ \angle 1$

$\displaystyle \angle ADC=\frac 1{2}\ \angle 2$

$\displaystyle \angle B+ \angle D=\frac 1{2}\ \angle 1+\frac 1{2}\ \angle 2$

$\displaystyle \angle B+ \angle D=\frac 1{2}\underbrace{(\angle 1+\angle 2)}_{\text{This is}\ 360^0}$

$\displaystyle \angle B+ \angle D=\frac 1{2}\times 360^0$

$\displaystyle \angle B+ \angle D=180^0$

OK?