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.
  

Draw a line through A parallel to XB, and extend XP to cross this line at Q.

Triangles APQ and BPX are similar (AQ||XB giving alternate angles) (1)

Thus:       tan/AXP / tan/XAP       = |AX|tan/AXQ / |AX|tan/XAB

                                                = |AQ| / |BX|    using (1)

                                                = |AP| / |BP|   (a fixed ratio).

Posted by Harry on 2013-12-17 17:17:20