A circle has diameter AB and a point P lies on AB between A and B.
A point X, distinct from A and B, lies on the circle.

Prove that tan(∠AXP)/tan(∠XAP) is constant for all values of X.
  

We can do a bit better than this. Construct a perpendicular to BX through P. Construct a line through X parallel to this perpendicular. Since X is on the circle, and the angle between the parallel and the perpendicular is 90 degrees, the parallel must pass through A.

Right triangles ABX and PBC are similar, and AB=2r.
2r/(BP)=AX/CP=BX/BC. [1]
Further angle AXP=CPX, and XAP=BPC (parallel lines) [2]
Now using Opp/Adj = tan, we have
CX/CP = tan AXP
BC/CP = tan XAP
Combining these: (CX/CP)/(BC/CP)=CX/BC
But since the ratio BC/BX is fixed for any compliant point P, and BC+CX = BX, the ratio CX/BX is also fixed. Hence for any given P, (tanAXP)/(tanXAP) is constant irrespective of X. Equally, for any compliant point X, the ratios BP/CP, and BP/BC are constant irrespective of P, since then the larger of the two similar triangles is fixed. 
More interestingly, from [1] it follows that BP*BX/BC = 2r , a constant for any given size of circle, irrespective of how P and X are chosen.

Edited on December 17, 2013, 10:43 pm

Posted by broll on 2013-12-17 22:42:21