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 \$\displaystyle \angle DAB\$ be x.
\$\displaystyle \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\$\displaystyle \angle DOB\$(angle sum of point O equals 360)
reflex\$\displaystyle \angle DOB\$= 360-2x
\$\displaystyle \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)
\$\displaystyle \angle DAB\$ + \$\displaystyle \angle DCB\$ = x + 180 - x
\$\displaystyle \angle DAB\$ + \$\displaystyle \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 \$\displaystyle \angle DAB~\&~\angle DCB\$ are opposite angles.
An inscribed angle is one-half the measure of its intercepted arc.
The union of the arcs \$\displaystyle \overbrace {DAB}^{}~\&~\overbrace {DCB}^{}\$ is the entire circle.
So what is the sum of the measures of the two opposite angles.