Prove that the 9-point circle bisects every segment connecting the orthocenter to a point on the circumcircle.
Thanks very much for your help!!!!
Another way of saying this is that the radius of the nine-point circle is half the radius of the circle circumscribing the triangle. Now, the 9 point circle can be thought of the circumscribing triangle of the midpoint of this triangle. Because if you have 3 non-colieanr points you can only draw a unique circle around them.Originally Posted by suedenation
Draw triangle ABC, then, mark off its midpoints XYZ. Draw triangle XYZ, we can show that XYZ is similar to ABC, by ratio of 2:1. Next, by the Extended law of sine, we know that for the big triangle and for the small triangle. But the sides are in proportion of 2:1 thus, but thet are similar thus, . From here we see .
Any problems? I explained it horrifically.