Let BB' be an altitude of right triangle ABC
(where B is the right angle).

Prove that the common chord of the circle with
center B and radius |BB'| and the circumcircle
of triangle ABC bisects BB'.

Let R be the intersection of the common chord, PQ, and BB’, with |BR| = a.
Let the circle with centre at B have radius r and B’D be a diameter.
Let BB’ cut the circumcircle of triangle ABC again at E. Since angle ABC is a right angle, AC is a diameter and BB’ = B’E = r.

Using the intersecting chord theorem in both circles:

|PR||RQ|  =                               |BR||RE|  =  |DR||RB’|


which gives                                a(2r - a)  =  (r + a)(r - a)

                                                2ar  - a²  =  r²  -  a²

                                                          a  =  r/2

proving that R is the mid-point of BB’.

Posted by Harry on 2010-09-24 22:44:18