**Problem:** Let be a triangle, is the circumcenter, is the nine point circle and is the Feuerbach triangle. The perpendicular to the line through the point intersects at the point (the points and lie on the same side of the line ) and similarly construct the points , . Prove that the lines , , are concurrent and the point of concurrency lies on

the line .

(See also: http://forumgeom.fau.edu/FG2005volume5/FG200521.pdf )

**My proof:**

Construct the circumcircle of triangle . Let be the midpoint of the arc which does not contain . Let be the midpoint of . intersects the A-altitude of triangle at , similar with . Let be the tangencies of and respectively.

Note that is the circumcenter of triangle then by Thales’s theorem:

But . Therefore or .

Let be the reflection of wrt then is a parallelogram. We get intersects at such that is the midpoint of . From my proof at the last corollary of Fontene’s theorem we can apply for the case and get lies on the circle with diameter . If is the projection of on then and so , or . But it is easy to see that are collinear then are collinear.

Let be the reflection of wrt . Construct the tangents of through , they intersect each other and make triangle . Note that and are the tangencies of A and A’-excircle with their sides so or . Similarly, we claim that and have . This means concur at their homothetic center.

Let be the tangencies of and be the tangencies of and .

From my alticle at here we obtain and A-altitude of triangle are concurrent and denote as their concurrency. Because then is the midpoint of .

Note that so by Thales’s theorem, must passes through the midpoint of , or .

Let be the projection of on be the projection of on . We have:

.

Note that then is the altitude of triangle , which follows that passes through the circumcenter of triangle . This means the orthocenter of triangle is the circumcenter of triangle

From and we get the homothetic center of two triangles and lies on . Our problem is solved.