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

  1. #1
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    Exclamation 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!
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  2. #2
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    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.
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