Circle Geometry

**xwrathbringerx** · January 21st 2009, 05:57 PM

Hi

AB and AC are equal chords of a circle. AM and BN are parallel chords through A and B respectively. Prove that AN is parallel to CM.

I've tried to prove angles NAM and AMC are equal but failed miserably. Could someone please help me with this question?

**red_dog** · January 22nd 2009, 01:24 AM

$m(\widehat{ABC})=m(\widehat{ACB})$
$m(\widehat{ABC})=m(\widehat{AMC})=\frac{m(arcAC)}{ 2}$
Then $m(\widehat{AMC})=m(\widehat{ACB})=\frac{m(arcAB)}{ 2}=\frac{m(arcAN)+m(arcNB)}{2}$

$m(\widehat{NAB})=\frac{m(arcNB)+m(arcBM)}{2}$

But $AM\parallel BN\Rightarrow m(arcAN)=m(arcBM)$

Then $m(\widehat{NAB})=\frac{m(arcAB)}{2}=m(\widehat{AMC })$

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