I was reading the the following theorem: Let $A,B$ be two points on the circumference of a circle. Let $C$ be a point outside the circle. Then $\angle BAC=\frac{1}{2}\widehat{AB}$. Is there some elementary way to prove it? Here is a picture with different points: http://jwilson.coe.uga.edu/emt668/EMAT6680.2003.fall/Nichols/6690/Webpage/Day%209.htm, the part "is formed by tangent and secant".
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@N. F. , your editing came messed up – Timbuc Sep 28 '14 at 11:47
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My edit required the yhmath package, so I nixed it. – N. F. Taussig Sep 28 '14 at 11:50
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Well, then undo it as it doesn't show correctly...never mind, I edited it already. – Timbuc Sep 28 '14 at 11:51
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The edit I made was to change \arc to \wideparen, which did not work. I undid it since readers will at least understand what geometrystudent meant by \arc. – N. F. Taussig Sep 28 '14 at 11:52
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@geometrystudent As given, your "theorem" is false. What is true is that $;\angle BAC=\frac12\widehat{BD};$ , where $;D;$ is the other point on the circle and on the line segment $;BC;$ – Timbuc Sep 28 '14 at 11:55
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I think you have understand the problem incorrectly. This is the case only when the segment $\overline{AC}$ is tangent to the circle. In this case we have $\measuredangle{BAC}=\frac{1}{2}\widetilde{AB}$, in which $\widetilde{AB}$, denotes the arc $AB$, and its proof is completely straightforward. I you draw the diameter passing from $A$, intersects the other side of the circle in $A'$.
$\measuredangle{BAC}=\frac{\pi}{2}-\measuredangle{BAA'}=\frac{\pi}{2}-\frac{1}{2}\widetilde{A'B}=\frac{\pi-\widetilde{A'B}}{2}=\frac{\widetilde{A'A}-\widetilde{A'B}}{2}=\frac{\widetilde{BA}}{2}$
CLAUDE
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