Let ABC be a triangle with /ABC < /ACB < 90°.
Let D be a point on side BC such that |AD| = |AC|.
The incircle of triangle ABC is tangent to sides
AC and BC at points P and Q respectively. Let J
be the incenter of triangle ABD.

Prove that line PQ bisects line segment CJ.

1. Construct the incircle of ABD (with incentre at J) as C1.
2. Construct the radius JR of C1 to its tangent point on BC.
3. Draw the circumcircle, C2, of CRJ. Naturally, since CRJ is a right angle, CJ is a diameter of C2.

4. Now it suffices to note that O lies at the midpoint of diameter CJ , so that JO = OC. (Equivalently, a circle drawn on O will intersect J,R,C, to the same effect.)

Edited on October 24, 2010, 8:31 am

Posted by broll on 2010-10-24 08:28:24