# Circle Geometry Proof - Opposite Angles of a Cyclic Quadrilateral are Supplementary

• May 17th 2009, 02:53 AM
xwrathbringerx
Circle Geometry Proof - Opposite Angles of a Cyclic Quadrilateral are Supplementary
Hi

I was wondering if anyone could please show me how to prove the theorem: opposite angles of a cyclic quadrilateral are supplementary. I know the way using:

Let $\angle DAB$ be x.
$\angle DOB$ = 2x (the angle at the centre of a circle is twice the angle at the circumference standing on the same arc DB)
360 = 2x + reflex $\angle DOB$(angle sum of point O equals 360)
reflex $\angle DOB$= 360-2x
$\angle DCB$= 180-x (the angle at the centre of a circle is twice the angle at the circumference standing on the same arc DB)
$\angle DAB$ + $\angle DCB$ = x + 180 - x
$\angle DAB$ + $\angle DCB$ = 180

However, I have not gotten up to the stage of proving (the angle at the centre of a circle is twice...). Could someone please show me an alternative way?

Thanx a lot!
• May 17th 2009, 04:12 AM
Plato
Suppose that ABCD is cyclic quadrilateral then $\angle DAB~\&~\angle DCB$ are opposite angles.
An inscribed angle is one-half the measure of its intercepted arc.
The union of the arcs $\overbrace {DAB}^{}~\&~\overbrace {DCB}^{}$ is the entire circle.
So what is the sum of the measures of the two opposite angles.