Thread: In quadrilateral, prove sum of angles AOB and COD =180...?

1. In quadrilateral, prove sum of angles AOB and COD =180...?

Let ABCD be quadrilateral inscribed in a circle with center O. Prove that sum of the angles AOB and COD = 180°.

2. Hello, fardeen_gen!

Is there a typo ... or omitted information?
. . The statement is not true . . .

Let $\displaystyle ABCD$ be a quadrilateral inscribed in a circle with center $\displaystyle O.$
Prove that: .$\displaystyle \angle AOB + \angle COD \:= \:180^o$
Counter-example . . .
Code:
          A   * * *   B
o           o
*   \       /   *
*     \     /     *
\   /
*        \ /        *
*         *O        *
*        / \        *
/   \
*     /     \     *
*   /       \   *
o           o
D   * * *   C

For the statement to be true,
. . $\displaystyle AC$ and $\displaystyle BD$ must be perpendicular diameters ... always!

3. I posted the question as it was given in the source. Thanks for the counter example.

4. Originally Posted by fardeen_gen
Let ABCD be quadrilateral inscribed in a circle with center O. Prove that sum of the angles AOB and COD = 180°.
Its not true, since angle AOB can be changed without altering angle COD their angle sum cannot be an invariant of a cyclic quadrilateral.

RonL

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AOBTis a cyclic quadrilateral

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