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

• Aug 17th 2008, 05:19 AM
fardeen_gen
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°.
• Aug 17th 2008, 05:49 AM
Soroban
Hello, fardeen_gen!

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

Quote:

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!

• Aug 17th 2008, 05:53 AM
fardeen_gen
I posted the question as it was given in the source. Thanks for the counter example.
• Aug 17th 2008, 05:56 AM
CaptainBlack
Quote:

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