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Orthocenters of Two Triangles Sharing Circumcenter and Base: What is it about?
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Copyright © 1996-2008 Alexander Bogomolny
Explanation
The applet attempts to suggest a curious fact and one possible way of explaining why it stands:
| The line joining the orthocenters of two triangles that share a base and the circumcenter, is equal and parallel to the line joining their apexes. |
This fact is a byproduct of a discussion on a remarkable line in a cyclic quadrilateral. Here we offer an independent demonstration.
The proof follows from the observation that in a triangle, the segment joining the orthocenter with a vertex is parallel to and twice as long as the perpendicular from the circumcenter to the opposite side. Two triangles that share a base and the circumcenter, also share that perpendicular. Thus the two segments that join their orthocenters with the apexes are equal and parallel and so form a pair of opposite sides of a parallelogram. The segments in question complete the parallelogram with another pair of equal and parallel sides.
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Now for a single triangle. Any triangle is homothetic in the centroid G to its medial triangle with the coefficient -1/2. The perpendicular bisector of a side serves as an altitude in the medial triangle. It follows that the circumcenter of the base triangle serves as the orthocenter of the medial triangle. Thus the homothety at hand maps one onto the other. (This is in fact the starting point for a discussion on the Euler line.) It also maps the apex C to the midpoint M of the base AB. If H is the orthocenter and O the circumcenter of ΔABC, CH is mapped onto OM and is therefore twice as long as the latter, and the two are parallel.
Copyright © 1996-2008 Alexander Bogomolny
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