# triangle orthocenter and circumcircle

• Mar 9th 2010, 10:26 AM
hollywood
triangle orthocenter and circumcircle
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.

Thanks!

I'm trying to help in this thread, but I got stuck.
http://www.mathhelpforum.com/math-he...dilations.html
• Mar 9th 2010, 11:05 AM
icemanfan
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.
• Mar 10th 2010, 09:53 AM
hollywood
Thanks, but I got the solution by another method - see the original thread:
http://www.mathhelpforum.com/math-he...dilations.html