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 . . .Let $\displaystyle ABCD$ be a quadrilateral inscribed in a circle with center $\displaystyle O.$
Prove that: .$\displaystyle \angle AOB + \angle COD \:= \:180^o$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!