Let A and B be different points on a circle with center O. With only a straightedge and a compass, can you construct a straight line through O meeting the segment AB at C, C strictly between A and B, and meeting the circle at D, so that C is between O and D and the segments AC and AD have the same length?

From triangle CAD:  <CAD + <ADC + < ACD = 180

From isosceles triangle ODA :  <ODA = <OAD

Combining these with the fact that  

  <ADC = <ODA  and  <OAC = <OAB  

gives

  <ODA = <OAD = <OAC + <CAD = <OAC + (180 - <ADC - <ACD)

       = <OAB + (180 - <ODA - <ACD)

                       or

  <OAB = 2 <ODA + <ACD - 180

Therefore,

  AD = AC  <=>  <ACD = <ADC = <ODA 

           <=>  <OAB = 3 (<ODA - 60)

                or

  AD = AC  <=>  <OAB/3 = <ODA - 60

Therefore, our construction is possible if and only if
<OAB can be trisected with a straightedge and compass.

Posted by Bractals on 2005-07-20 16:23:16