Let O be the center of the circle. If you can show that OQB is a right angle, then triangles OQB and OQC are congruent, which means QB is the same length as QC.
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.
Thanks, but I got the solution by another method - see the original thread: