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