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 $\displaystyle 2R=\frac{a}{\sin \alpha}$ for the big triangle and $\displaystyle 2r=\frac{x}{\sin X}$ for the small triangle. But the sides are in proportion of 2:1 thus, $\displaystyle 2x=a$ but thet are similar thus, $\displaystyle \sin \alpha= \sin X$. From here we see $\displaystyle 2r=R$.
Any problems? I explained it horrifically.