Siamese Triangles II: What Is This about?
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Copyright © 1996-2012 Alexander Bogomolny
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Given ΔABD and a point C on its base line BD. The triangles ABC and ACD share a side AC and a base line. In each of the latter two triangles, consider in succession the centroid, the circumcenter, and the orthocenter. Through those points draw parallels to the free side of the other triangle. I claim that in each case the parallels intersect on the base line BD.
I'll use the dynamic approach that worked so well on another occasion. Let's fix ΔABD, but allow point C glide over BD. With each position of C, we associate two lines: one parallel to AB, the other to AD passing through the denominationally corresponding points of triangles ACD and ABC. For the centroid, the circumcenter and the orthocenter it is quite clear that if the lines intersect on BD for one position of C, then the same is true for all other positions as well. (What is lost or gained on the left from C, is gained or lost to the right from C. You'll need an accurate drawing to see why.)
Thus the problem is reduced to finding a position for C, for which the claim is obvious. For the centroid we may take C to be the midpoint of BD. The lines in question then will be the midlines of ΔABD drawn from C. For the circumcenter and the orthocenter C could be taken to be the foot of the altitude from A. Then both ΔABC and ΔACD are right. Their circumcenters lie on AB and AD, respectively, such that the lines in question are again the midlines of ΔABD. As regard the lines through the orthocenter, the situation is even simpler, since in this case, the orthocenters of both triangles ABC and ACD lie at C.
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Copyright © 1996-2012 Alexander Bogomolny
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