Here is my problem:
Given an arbitrary triangle ABC, let A' denote the diametral opposite of A in the circumcircle of triangle ABC. If H is the orthocenter of triangle ABC, show that segments HA' and BC bisect each other.
I tried to proceed by proving triangle CQA' and triangle BQH (where Q is the intersection of HA' and BC) are congruent, which would give me the result I need. But I couldn't see how to get there.
I'm trying to help in this thread, but I got stuck.