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March 19th, 2012, 11:53 PM   #1
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Circle Proof

[attachment=0:v0wrhbh4]circleproof2.jpg[/attachment:v0wrhbh4]
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 March 20th, 2012, 05:35 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Circle Proof That link requires a registration/login. Please attach the image instead.
March 20th, 2012, 03:36 PM   #3
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Re: Circle Proof

[color=#000000][attachment=0:q4f0iqds]pap.png[/attachment:q4f0iqds]

Here is a hint: (hope is ok since the time has passed for me)

Draw the line segment ab, then the perpendicular line from the point b to ax and to rP respectively. Now since you are given that the red and green colored angles are equal you can easily prove that the triangles ${\triangle}bDx$ and ${\triangle}xPb$ are equal. Now expand the line segments ab and xb as shown in the figure and draw the parallel line to ax which passes from point b. Look close and you can prove that the angle $\angle xab$ is equal to $\angle ax b$ which will lead you to the equality of the triangles $\triangle aDb$ and $\triangle xDb$. So we deduce that $ab=xb\Rightarrow \overset{\frown}{ab}=\overset{\frown}{xb}$ and so b is the midpoint of $\overset{\frown}{abx}$.[/color]
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 March 20th, 2012, 08:01 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1808 It would be easier to use the alternate segment theorem. The minor arc/chord BX subtends angle BAX, that equals angle BXP. The minor arc/chord AX subtends an acute angle that equals angle AXP. Since angle BXP is half angle AXP, the minor arc BX is half the arc ABX, so B is the midpoint of the arc ABX. The third step uses a theorem that the acute angle subtended by a minor arc is proportional to the length of the arc.
 March 20th, 2012, 09:06 PM #5 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Circle Proof Here is another way to look at it: By alternate segment theorem, we can say that$\angle BXP= \angle BAX$. Let $\angle BXP= \angle BAX=\alpha$ Since BX bisects angle AXP, we have $\angle AXB=\angle BXP=\alpha$ Now, consider the triangle ABX. We see that $\angle XAB=\angle AXB=\alpha$, we can say triangle ABX is an isosceles triangle with AB=BX. But congruent chords have congruent arcs, i.e. arc AB=arc BX, therefore it must be true that B is the midpoint of the arc ABX.

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