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Math Help - sum of opposite angles in any cyclic quadrilateral is 180

  1. #1
    Member u2_wa's Avatar
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    Lightbulb 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.
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  2. #2
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    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 ABCD.

    . . \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]:

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

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

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

    . . . . . . . . . =\;180^o



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

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  3. #3
    Member u2_wa's Avatar
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    Sir, kindly tell me what do you mean by 0.5 arc (BCD)?
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  4. #4
    Member great_math's Avatar
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    Quote Originally Posted by u2_wa View Post
    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 O



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

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

    Adding we have ,
    \angle B+ \angle D=\frac 1{2}\ \angle 1+\frac 1{2}\ \angle 2

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

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

    \angle B+ \angle D=180^0


    OK?
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