Thread: 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°.

Hello, fardeen_gen!

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

Let be a quadrilateral inscribed in a circle with center
Prove that: .

Counter-example . . .

Code:

          A   * * *   B
          o           o
        *   \       /   *
       *     \     /     *
              \   /
      *        \ /        *
      *         *O        *
      *        / \        *
              /   \
       *     /     \     *
        *   /       \   *
          o           o
          D   * * *   C

For the statement to be true,
. . and must be perpendicular diameters ... always!

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

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