March 19th, 2012, 11:53 PM  #1 
Newbie Joined: Mar 2012 Posts: 2 Thanks: 0  Circle Proof
Please help solve it. I'm stuck and this is my last problem. [attachment=0:v0wrhbh4]circleproof2.jpg[/attachment:v0wrhbh4] 
March 20th, 2012, 05:35 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Re: Circle Proof
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March 20th, 2012, 03:36 PM  #3 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  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 and 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 is equal to which will lead you to the equality of the triangles and . So we deduce that and so b is the midpoint of .[/color] 
March 20th, 2012, 08:01 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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. Let Since BX bisects angle AXP, we have Now, consider the triangle ABX. We see that , 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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