[Here's a synthetic geometry problem passed to me by Bill Blair in
summer 1996, together with a solution by him. As I understand it, this
was a tough homework problem unsolved by a young friend of his. -- djr]


Suppose given a circle with a chord TU and two points C and  A on the circle 
lying on the same side of the chord (so  T C A U  is an arc of the circle).
Perform the following constructions.

Let  M  be the midpoint of the segment TU. 
Draw lines CMD and AMB through  M, meeting the circle at two new points
B and D. Then draw segments CB and AD. Let P and Q be the points where
these segments cross the chord TU.

Prove that MP = MQ.
                                  C o******                   
                               ****/\      ****** A             
                            ***   /  |         __o**           
                         ***    _/    \     __/  | ***         
                       ***   P /       | __/     |    **       
                    T o*******o********o*********o******o U      
                     **     _/    __/   \        \ Q     **    
                    **    _/   __/     M |        |       *    
                    *    /  __/           \       |        *   
                   *   _/__/               |      |        *   
                   *  /_/                   \     \        **  
                   * //                      |     |       * 
                   o/                         \    |       *
                 B  *                          |   |       *   
                    **                          \   \     *    
                     **                          |  |    **    
                      **                          \ |   *      
                       ***                         || **       
                         ***                       *o*         
                            ***                  ***            
                               *****       ******    D          
                                    *******                   


SPOILER follows (taken verbatim from Blair, but reformatted for email.)


Construct perpendiculars PS and PR to AB and CD respectively, and QW
and QV to CD and AB respectively.

Since triangle MRP is similar to MWQ we have MP/MQ = PR/QW.
Since triangle MSP is similar to MVQ we have MP/MQ = PS/QV.

Since angle BCD is congruent to BAD we have that
	triangle CRP is similar to AVQ and thus PR/QV=CP/AQ.
Since angle CBA is congruent to CDA we have that
	triangle BSP is similar to DWQ and thus PS/QW=BP/DQ.

Hence
	MP^2/MQ^2 =
	= PR/QW . PS/QV
	= PR/QV . PS/QW 
	= (CP.BP)/(AQ.DQ)
	= (TP.PU)/(UQ.QT)
	= (MT-MP).(MT+MP) / (MU-UQ).(MU+MQ)
	= (MT^2-MP^2)/(MU^2-MQ^2)
	= (MT^2-MP^2)/(MT^2-MQ^2)

Thus MP^2(MT^2-MQ^2) = MQ^2(MT^2-MP^2) and so 
	MP^2MT^2 - MP^2MQ^2 = MQ^2MT^2-MP^2MQ^2
and hence MP^2MT^2 = MQ^2MT^2. Therefore MP^2=MQ^2 and so MP=MQ.


Remark: The feet  S, V  of the constructed perpendiculars PS and QV
might not fall in the interior of the segment  AB as shown, bu this
doesn't matter. Similarly the feet R,W of the perpendiculars PR and
QW might not fall in the interior of the segment CD as shown.