Let PQ be a diameter of a circle,
     A be a point on line PQ such that P lies between A and Q,
     T be a point on the circle such that line AT is tangent to the circle,
     B be the point on line QT such that line BP is perpendicular to line PQ, and
     C be the point on line PT such that line CQ is perpendicular to line PQ.

Prove that points A, B, and C are collinear.

Let R and S be the points of intersection of AT with PB and QC respectively.

Equal tangents and radii make OQST a kite, with OS perpendicular to TQ and therefore parallel to PC (since angle PTQ is a right angle). Since O is the mid-point of PQ, S must be the mid-point of QC (triangles QOS, QPC are similar, AAA).

Using exactly the same argument based on the kite OPRT, it follows that R is the mid-point of PB.

Since the parallel line segments PB and QC are bisected by the line RS, it follows (using similar triangles) that RS will bisect all other parallel line segments formed between the lines BC and PQ. Therefore CB, SR and QP are concurrent at A, proving that A, B and C are collinear.

Edited on December 13, 2009, 12:02 pm

Posted by Harry on 2009-12-13 11:59:59